/^ ^,(ru. ELEMENTS INTEGRAL CALCULUS KEY TO THE SOLUTION OF DIFFERENTIAL EQUATIONS. BY WILLIAM ELWOOD BYERLY, Ph.D., 5, PKOFESSOK OF MATHEMATICS IN HARVARD UNIVERSITY. BOSTON COLLEGE PHYSICS DEPT. BOSTON: PUBLISHED BY GINN, HEATH, & CO. 1881. ELEMENTS INTEGRAL CALCULUS KEY TO THE SOLUTION OF DIFFERENTIAL EQUATIONS. BY WILLIAM ELWOOD BYERLY, Ph.D., 5, PROFESSOR OF MATHEMATICS IN HARVARD UNIVERSITY. BOSTON COLLEGE PHYSICS DEPT. BOSTON: PUBLISHED BY GINN, HEATH, & CO. 1881. Enterea according to Act of Congress, in the year 1S81, by William Elavood Byekly, in the office of the Librarian of Congress, at Washington. GiNN, Heath, & Co. J. S. CusHiNG, Printer, 16 Hawley Street, Boston. 260413 PREFACE. The following volume is a sequel to my treatise on the Differential Calculus, and, like that, is written as a text-book. The last chapter, however, a Key to the Solution of Differential Equations, maj^ prove of service to working mathematicians. I have used freely the works of Bertrand, Benjamin Peirce, Todlmnter, and Boole ; and I am much indebted to Professor J. M. Peirce for criticisms and suggestions. I refer constantly to my work on the Differential Calculus as Volume I. ; and for the sake of convenience I have added Chapter V. of that book, which treats of Integration, as an appendix to the present volume. W. E. BYERLY. Cambridge, 1881. ANALYTICAL TABLE OF CONTENTS. CHAPTER I. SYMBOLS OF OPERATION. Article. Page. 1. FivnctioViBl symbols regarded as symbols of ojjeration 1 2. Compoimd function ; compound operation 1 3. Commutative or relatively free operations 1 4. Distributive or linear operations 2 5. The compounds of distributive operations are distributive ... 2 6. Symbolic exponents 2 7. The law of indices 2 8. The interpretation of a zero exponent 3 9. The interpretation of a negative exponent . 3 10. When operations are commutative and distributive, the sym- bols which represent them may be combined as if they were algebraic quantities 3 CHAPTER II. EVIAGESTAKIES. 11. Usual definition of an imaginary. Imaginaries first forced upon our attention in connection with quadratic equations .... 5 12. Treatment of imaginaries purely arbitrary and conventional . . (J 13. V^ defined as a symbol of operation fl 14. The rules in accordance with which the symbol V — 1 is used. V^ distributive and commutative with symbols of quantity . 7 15. Interpretation of powers of V^l 7 16. Imaginary roots of a quadratic 8 17. Typical form of an imaginary 8 18. Geometrical representation of an imaginary. Beals and 2mre imaginaries. An interpretation of the operation V^^ ... 8 19. The sum, the product, and the quotient of two imaginaries, a + b V^ and c + d V^, are imaginaries of the typical form 10 vi INTEGRAL CALCULUS. Article. Page. 20. Second typical form r (cos f + V— 1 sin (p). Modulus and argii- ment. Examples 10 21. The modulus of the sum of two imaginaries is never greater than the sum of their moduli 11 22. Modulus and argument of the product of imaginaries 12 23. Modulus and argument of the quotient of two imaginaries . . 13 24. Modulus and argument of a power of an imaginary 13 25. Modulus and argument of a root of an imaginary. Example . 14 26. Relation between the n nth roots of a real or an imaginary , . 14 27. The imaginary roots of 1 and — 1. Examples 15 28. Conjugate imaginaries. Examples 17 29. Transcendental functions of an imaginary variable best defined by the aid of series 17 30. Couvergency of a series containing imaginary terms 18 31. Exponential functions of an imaginary. Definition of e" where z is imaginary 19 32. The law of indices holds for imaginary exponentials. Example 20 33. Logarithmic functions of an imaginary. Definition of log z. Log 2; a pe?'i;ofZtc function. Example 21 34. Trigonometric functions of an imaginary. Definition of sin z and cos z. Example 22 35. Sin z and cos z expressed in exponential form. The fundamen- tal formulas of Trigonometry hold for imaginaries as well as for reals. Examples 22 36. Differentiation of Functions of Imaginary Variables. The de- rivative of a function of an imaginary is in general indetermi- nate 24 37. In difi"erentiating. we may treat the V—i like a constant factor. Example. Two forms of the differential of the independent variable , 24 38. Differentiation of a power of z. Example 25 39. Differentiation of e^. Example 26 40. Differentiation of log z 26 41. Diflerentiatiou of sir^g and coss 26 42. Formulas for direct integration (I., Art. 74) hold when x is imaginary 27 43. Hyperholic Functions 27 44. Examples. Properties of Hyperbolic Functions 28 45. Differentiation of Hyperbolic Functions 28. 46. Anti-hyperbolic functions. Examples 28 47. Anti-hyperbolic functions expressed as logarithms 29 48. Formulas for the direct integration of some irrational forms . 30 TABLE OF CONTENTS. vii CHAPTER III. GENERAL METHODS OF INTEGRATINCr. Article. Page. 49. Integral regarded as the inverse of a differential 32 50. If /x is any function whatever of x, fx.dx has an integral, and but one, except for the presence of an arbitrary constant . . 32 51. A definite integral contains no arbitrary constant, and is a func- tion of the values between which the sum is taken. Exam- ples 33 52. Definite integral of a disco Hf Ml i(o?fS function 33 53. Formulas for direct integration 34 54. Integration by substitution. Examples 36 55. Integration by parts. Examples. Miscellaneous examples in integration 37 CHAPTER IV. EATIONAJL FRACTIONS. 56. Integration of a rational algebraic polynomial. Rational frac- tions, proper and improper 40 57. Every proper rational fraction can be reduced to a sum of sim- pler fractions with constant numerators 40 58. Determination of the numerators of the partial fractions by indirect methods. Examples 42 59. Direct determination of the numerators of the partial fractions 43 60. Illustrative examples 45 61. Illustrative example 46 62. Integration of the partial fractions 48 63. Treatment of imaginary values which may occur in the partial fractions. Examples 49 CHAPTER V. REDUCTION FORMULAS. 64. Formulas for raising or lowering the exponents in the form x'^-'^{a-\-hx'^)Pdx 52 65. Consideration of special cases. Examples 54 Viii USTTEGEAL CALCULUS. CHAPTER VL IRRATIONAL FORMS. Article. ^^Se 66. lategration of the form f(x, V a + bx)dx. Examples 56 67. lutegrationof the form/(x, Vc+ \/a + to)f?x. Examples . . 57 68. Integration of the form /(x, Va + ta + cx'Of^a; 57 69. Illustrative example. Examples 59 70. Integration of the form /fx,A/^^-t-V^3;. Example 61 \ ^Ix+mJ 71. Application of the Beduction Formulas of Chapter V. to irra- tional forms. Examples 61 72. A function rendered irrational through the presence under the radical sign of a polynomial of higher degree than the second cannot ordinarily be integrated. Elliptic Integrals . 62 CHAPTER VII. TRANSCENDENTAL FUNCTIONS. . 73. Use of the method of Integration by Farts. Examples .... 63 74. Reduction Formulas for sin" X and cos" X. Examples 64 75. Integration of (sin-ix)«(:?x. Examples 65 76. Use of the method of Integration by Substitution 66 77. Integration of sin™ X COS" x.dx. Examples 67 CHAPTER VIII. r)EFIN^TE INTEGRALS. 78. Computation of a definite integral as the limit of a sum .... 69 79. Illustrative example. Examples 70 80. Usual method of obtaining the value of a definite integral. Examples • 70 81. Application of reduction formulas to definite integrals. Ex- amples 71 82. The Gamma Function 73 83. The x>rincipal value of the definite integral of a discontinuous function. Example 74 84. Difi'erentiation and integration of a definite integral when the limits of integration are constant 74 TABLE OF CONTENTS. IX Article. Page. 85. Simplification of a complicated form by differentiation under the sign of definite integration. Example 76 86. Simplification of a complicated form by integration under the sign of definite integration. Examples 77 87. Differentiation of a definite integral when the limits of integra- tion are not constant. Example 77 88. Definite integrals taken between imaginary limits 78 CHAPTEK IX. LENGTHS OF CURVES. 89. Formulas for sin t and cos t in terms of the length of the arc . 79 90. The equation of the Catenary obtained. Example 79 91. The equation of the Tractrix. Examples 81 92. Length of an arc in rectangular coordinates 83 93. Length of the arc of the C?/do«cL Example 84 94. Another method of rectifying the Cycloid 85 95. Hecti&ca.tion of the Epicycloid. Examples 85 96. Arc of the Ellipse. Auxiliary angle. Example 86 97. Length of an arc. Polar coordinates . 87 98. Equation of the Logarithmic Spiral 87 99. Eectification of the Logarithmic Spiral. Examples 88 100. Rectification of the Cardioide 88 101. Involutes. Illustrative example. Example 89 102. The involute of the Cycloid. Example 91 103. Intrinsic equation of a curve. Example 92 104. Intrinsic equation of the Epicycloid. Example 93 105. Intrinsic equation of the Logarithmic Spiral 94 106. Method of obtaining the intrinsic equation from the equation in rectangular coordinates. Examples 94 107. Intrinsic equation of an evolute 96 108. Illustrative examples. Examples 96 109. The evolute of an Epicycloid. Example 97 110. The intrinsic equation of an involute. Illustrative examples . 9§ 111. Limiting form approached by an involute of an involute . . . 99 112. Method of obtaining the equation in rectangular coordinates from the intrinsic equation. Illustrative example 100 113. Rectification of Curves in Space. Examples 101 INTEGEAL CALCULUS. CHAPTER X. AREAS. Article. Page. 114. Areas expressed as definite integrals, rectangular coordi- nates. Examples 103 115. Areas expressed as definite integrals, polar coordinates . . .104 116. Area between the catenary and the axis 104 117. Area between the tractrix and the axis. Example 104 118. Area between a curve and its asymptote. Examples .... 105 119. Area of circle obtained by the aid of an auxiliary angle. Ex- amples lOG 120. Area between two curves (rect. coor.). Examples 107 121. Areas in Polar Coordinates. Examples 108 122. Problems in areas can often be simplified by transformation of coordinates. Examples Ill 123. Area between a curve and its evolute." Examples Ill 124. Holditch's Theorem. Examples 112 125. Areas by a double integration (rect. coor.) 114 126. Illustrative examples. Examples 115 127. Areas by a double integration (polar coor.). Example .... 116 CHAPTER XI. AREAS OF SURFACES. 128. Area of a surface of revolution (rect. coor.). Example .... 118 129. Illustrative examples. Examples 119 180. Area of a surface of revolution by transformation of coordi- nates. Example 120 131. Kvedi of a, surface of revolution (j^o\&Y coor.'). Examples . . .122 132. Area of any siirface by a double integration 122 133. Illustrative example. Examples 125 ^CHAPTER XIL VOLUMES. 134. Volume by a single integration. Example 128 135. Volume of a conoid. Examples 129 136. Volume of an ellipsoid. Examples 130 137. Volume of a solid of revolution. Single integration. Exam- ples 131 138. Volume of a solid of revolution. Double integration. Exam- ples 132 TABLE OF CONTENTS. xi Article. Page. 139. Yolnme of a solid of revolution. Polar formula. Example. .134 140. Volume of any solid. Triple iutegratiou. Rectangular coor- dinates. Examples 135 141. Volume of miy solid. Triple integration. Polar coordinates. Example 138 CHAPTER XIII. CENTRES OF GRAVITY. 142. Centre of Gravity defined 139 143. General formulas for the coordinates of the Centre of Gravity of any mass. Example 139 144. Centre of Gravity of a homogeneous body 141 145. Centre of Gravity of a j9Zrt we area. Examples 141 146. Centre of Gravity of a homogeneous solid of revolution. Ex- amples • 144 147. Centre of Gravity of an arc ; of a surface of revolution. Exam- ples 146 148. Properties of Guldin. Examples 147 CHAPTER XIV. MEAN VALUE AND PROBABILITY. 149. References 149 150. Mean value of a contimiously varying quantity. The mean dis- tance of all the points of the circumferences of a circle from a fixed point on the circumference. The mean distance of points on the surface of a circle from a fixed point on the cii'cumference. The mea^i distance between two points within a given circle 149 151. Problems in the application of the Integral Calculus to pro&a- bilities. Random straight lines. Examples 151 CHAPTER XV. KEY TO THE SOLUTION OF DIFFERENTIAL EQUATIONS. 152. Description of Key 158 153. Definition of the terms differential equation, order, degree, linear, general sohition or complete primitive, singular solu- tion. 154. Examples illustrating the use of tfie Key 159 Key 1G3 Examples under Key 180 INTEGEAL CALCULUS. CHAPTER I. SYMBOLS OF OPEKATIOIST. 1. It is often convenient to regard a functional symbol as indicating an operation to he ])er formed upon the expression tvhich is ivritten after the symbol. From this point of view the s^Tnbol is called a symbol of operation, and the expression writ- ten after the s3'mbol is called the subject of the operation. Thus the S3'mbol D^ in D^{x-y) indicates that the operation of differentiating with respect to a; is to be performed upon the subject (x^y). 2. If the result of one operation is taken as the subject of a second, there is formed what is called a compound function. Thus log sin?; is a compound function, and we may speak of the taking of the log sin as a compound operation. 3. When two operations are so related that the compound operation, in which the result of performing the first on any subject is taken as the subject of the second, leads to the same result as the compound operation, in which the result of per- forming the second on the same subject is taken as the subject of the first, the two operations are commutative or relatively free. Or to formulate ; if fFu^Ffu, the operations indicated by / and F are commutative. 2 INTEGRAL CALCULUS. [Art. 4. For example ; the operations of partial differentiation with respect to two independent variables x and y are commutative, for we know that D,DyU = DyD,u. (I. Art. 197). The operations of taking the sine and of taking the logarithm are not commutative, for log sin w is not equal to sin log m. 4. If f{u±v)=fu±fv where u and v are any subjects, the operation /is distributive or linear. The operation indicated b}' d and the operation indicated by D^ are distributive, for we know that . d{u ±v) = du ±dv, and that D^{u ±v) = D^u ± D^v. The operation sin is not distributive, for sin(i(H-'y) is not equal to sinu + sin-y. 5. The com^jounds of distributive operations are distributive. ■ Let / and F indicate distributive operations, then /i^ will be distributive ; for F{u ±v) = Fu ± Fv, therefore fF{u ±v)= f{Fu ± Fv) = fFu ± fFv. 6. The repetition of any operation is indicated by writing an exponent., equal to the number of times the op>eratio7i is ^je?-- formed^ after the symbol of the operation. Thus log^a; means log log log .'c ; d^u means dddui In the single case of the trigonometric functions a different use of the exponent is sanctioned by custom, and sin^u means (sin^t)^ and not sin sinw. 7. If m and n are ivhole numbers it is easily proved that Chap. I.] SYMBOLS OF OPERATION. 3 This formula is assumed for all values of m and n, and nega- tive and fractional exponents are interpreted by its aid. It is called the laio of indices. 8. To find what interpretation must be given to a zero ex- ^ ' m = in the formula of Art. 7. /o/«t^=/o + »'«=/"«, or, denoting /""it hj v, f°v = v. That is ; a symbol of operation with the exponent zero has no effect on the subject, and may be regarded as multiplying it b}' unit}'. 9. To interpret a negative exponent, let m= —n in the formula of Art. 7. /- ''f"-u^f-'' + ''u=fu = u. If we call f"u = V, then f-''v = u. If II = 1 we get f~^fu=ic, and the exponent —1 indicates what we have called the anti- function of fu. (I. Art. 72.) The exponent — 1 is used in this sense even with trigonometric functions. 10. "Wlien two operations are commutative and distributive, the symbols which represent them ma}' be combined precisely as if they were algebraic quantities. For they obey the laws, a (m -j- n) = am + an, am = ma, on which all the operations of arithmetic and algebra are founded. 4 INTEGEAL, CALCFLUS. [Art. 10. For example ; if the operation {D^ + Dy) is to be* performed n times in succession on a subject tt, we can expand (Z>^ + Z)j,)'* precisel}' as if it were a binominal, and tlien perform on u the operations indicated by tlie expanded expression. Chap. II.] mAGIXAKIES. CHAPTER II. IMAGINAEIES. 11. An imaginary is usually defined in algebra as the indi- cated even root of a negative quantity, and although it is clear that there can he no quantity that raised to an even power will be negative, the assumption is made that an imaginar}' can be treated Wko. any algebraic quantit}'. Imaginaries are first forced upon our notice in connection with the subject of quadratic equations. Considering the t3'pical quadratic „ X' -\- ax -\- = 0, we find that it has two roots, and that these roots possess cer- tain important properties. For example ; their sum is —a and their product is h. "We are led to the conclusion that every quadratic has two roots whose sum and whose product are simply related to the coefficients of the equation. On trial, however, we find that there are quadratics having but one root, and quadratics having no root. For example ; if we solve the equation cc2-2a; + l = 0, we find that the onl}' value of x which will satisfy it is unity ; and if we attempt to solve ' x' — 2x + 2 = 0, we find that there is no value of x which will satisfy- the equation. As these results are apparently inconsistent with the conclu- sion to which we were led on soMng the general equation, we naturall}' endeavor to reconcile them with it. The difficulty in the case of the equation which has but one 6 INTEGRAL CALCULUS. [Art. 12. root is easily overcome by regarding it as having two equal roots. Thus we can saj* that each of the two roots of the equation ic^ - 2 .T + 1 = is equal to 1 ; and there is a decided advantage in looking at the question from this point of view, for the roots of tliis equation will possess the same properties as those of a quadratic having unequal roots. The sum of the roots 1 and 1 is minus the co- efficient of X in the equation, and then' product is the constant term. To overcome the difficulty presented by the equation which has no root we are driven to the conception of imaginaries. 12. An imaginary is not a quantity, and the treatment of imaginaries is purely arbitrary and conventional. We begin by lading down a few arbitrar}'^ rules for our imaginary expressions to obe}', which must not involve an}' contradiction ; and we must perform all our operations upon imaginaries, and must interpret all our results by the aid of these rules. Since imaginaries occur as roots of equations, thej^ bear a close analog}' with ordinary algebraic quantities, and they have to be subjected to the same opei'ations as ordinary quantities ; thei'e- fore our rules ought to be so chosen that the results may be comparable with the results obtained when we are dealing with real quantities. 13. By adopting the convention that V— a^ = a V — 1, where a is supposed to be real, we can reduce all our imaginary algebraic expi-essions to forms where V — 1 is the only peculiar symbol. This s}'mbol V — 1 we shall define and use as the sym- bol of some operation, at present unknown, the repetition ofivhich has the effect of changing the sign of the subject of the operation. Thus in a'\l —\ the symbol V — 1 indicates that an operation is performed upon a which, if repeated, will change the sign of o. That is, a(V — 1)^= —a. Chap. II.] IMAGINAEIES. 7 From this point of view it would be more natural to write the symbol before instead of after the subject on which it operates, (V — l)a instead of aV — 1, and this is sometimes done; but as the usage of mathematicians i^ overwhelmingi}^ in favor of the second form, we shall emplo}' it, merel}^ as a matter of con- venience, and remembering that a is the subject and the V — 1 the symbol of operation. 14. The rules in accordance with which we shall use our new s^'mbol are, first, In other words, the operation indicated by V — 1 is to be dis- tributive (Art. 4) ; and second, aV"^=(V^l)a, [2] or our sj'mbol is to be commutative with the sj'mbols of quantity (Art. 3) . These two conventions will enable us to use our S3'mbol in algebraic operations precisel}' as if it were a quantity (Art. 10). When no coefficient is written before V — 1 the coefficient 1 will be understood, or unity will be regarded as the subject of the operation. 15. Let us see what interpretation we can get for powers of V — 1 ; that is, for repetitions of the operation indicated by the symbol. (^^1)0=1 (Art. 8), (V^l)i=V^3, ( V^l ) ' = - 1 , by definition (Art. 13), (V^ri)3= (V31)2V^ri = -V^I, by definition, (V31)4=_(V^1)2 =1, (V^l)5=iV^l =v^=n, (V^r-i)6^(V^^)^ =-1, and so on, the values V— 1, —1, — V— 1, 1, occumng in 8 INTEGRAL CALCULUS. [Akt. 1G. cj'cles of four. "We can formulate this as follows ; let n be zero or anj'^ positive whole number, then, 16. The definition we have given for the square root of a negative quantity, and the rules we have adopted concerning its use, enable us to remove entirel}^ the difficult}' felt in dealing with a quadratic which does not have real roots. Take the equation a;2_2a; + 5 = 0. (1) Solving by the usual method, we get x=l ± V^^ ; V^^ = 27"^, by Art. 13 [1] ; hence iK= 1 + 2V^ or 1 — 2 V"^. On substituting these results in turn in the equation (1), per- forming the operations hj the aid of our conventions (Art. 14 [1] and {2]), ancl interpreting (V — l)^ b}' Art. 15, we find that they both satisfy the equation, and that they can therefore be regarded as entirel}- analogous to real roots. "We find, too, that their sum is 2 and that their product is 5, and consequentl}' that thej' bear the same relations to the coefficients of the equation as real roots. 17. An imaginary root of a quadratic can alwaj's be reduced to the form a-j-b V— 1 where a and b are real, and this is taken as the general type of an imaginary ; and part of our work will be to show that when We subjecfimaginaries to .the ordinary functional operations, all our results are reducible to this tj'pical form. Chap. II.] IMAGINAEIES. 9 18. We have defined V — 1 as the symbol of an operation whose repetition changes the sign of the subject. Several different interpretations of this operation have been suggested, and the following one, in which every imaginary is graphicall}" represented by the position of a point in a plane, is commonly adopted, and is found exceedingly useful in suggest- ing and interpreting relations between different imaginaries and between imaginaries and reals. In the Calculus of Imaginaries, a + & V — 1 is taken as the general sj'mbol of quantity. If b is equal to zero, a-\-b V — 1 reduces to a, and is real; if a is equal to zero, a + & V — 1 re- duces to b V — 1 , and is called a piire imaginary. a + 5 V — 1 is represented by the position of a point referred to a pair of rectangular axes, as in analytic geometry, a being taken as the abscissa of the point and b as its ordinate. Thus in the figure the position of the point P represents the imaginary a-\-b V— 1 . If 6 = 0. and our quantity' is real, P will lie on the axis of X, which on that account is called the axis of reals; if a = 0, and we have a j^ure imaginary, P will lie on the axis of Y, which is called the axis of pure imaginaries. Since a and a V — 1 are represented by points equally distant from the origin, and lying on the axis of reals and the axis of pure imaginaries respectively, we maj^ regard the operation indicated by V — 1 as causing the point representing the subject of the operation to rotate about the origin through an angle of 90°. A repetition of the operation ought to cause the point to rotate 90° further, and it does ; for a(V— 1)^= —a, and is represented b}^ a point at the same distance from the 10 ESTTEGEAL CALCULUS. [Art. 19. origin as a, and Ij-ing on the opposite side of the origin ; again repeat the operation, and the point has rotated 90° further ; repeat again, and the point has rotated through 360°. We see, then, that if the subject is a real or a j^i'^re imaginary the effect of performing on it the operation indicated by V — 1 is to rotate it about the origin through the angle 90°. We shall see later that even when the subject is neither a real nor a pure imaginary, the effect of operating on it Anth V — 1 is still to produce the rotation just described. 19. The sum, the product, and the quotient of any two imagi- naries, a + & V — 1 and c + cZ V — 1 , are imaginaries of the typi- cal form. a + &V^n" + c + dV^ =a + c + (6 + cOV^^. [1] (a + W^) (c + d V^) =ac- bd + (be + ad)V^. [2] a+b-\/^A __ (a+h-\r~i) (c— dV"^) _ ac+bd+(bc—ad)^/'^ c+d\/^l (c+dV^) (c-dV^) c^ + d' ac -^bd be — ad , ■ c^ + dr ' c^ + d^ ^~^- ^^^ All these results are of the form A + jBV— 1. 20. The graphical representation we have suggested for imaginaries suggests a second tj'pical form for an imaginary. ■ Given the imaginary cb + ?/V — 1, let the x>olar coordinates of the point P which represents x-{-y'\/ — 1 be r and (^. r is called the modidus and ?/ = r sin ^ ;} d)=tan-i^. ^ X x + y^—l = rcos<^+ (V — l)rsin(^ = r{coscJ3 +V — l.sin^), [1] [2] [3] where the imaginary is expressed in terms of its modulus and argument. The value of r given b}'' our formulas [2] is ambiguous in sign ; and ^ ma}- have any one of an iiifinite number of values differing by multiples of tt. In practice we alwa3'S take the positive value of r, and a value of <^ which will bring the point in question into the right quadrant. In the case of any given imaginar}^ then, r can have but one value, while 4> ma}' have any one of an infinite number of values differing by 2 tt. Examples. (1) Find the modulus and argument of 1 ; of V — 1 ; of — 4 ; of — 2V — 1 ; ofS + SV — 1; of2+4V— 1; and express each of these quantities in the form ?'(coS(^ +V — 1. sin^). (2) Show that every positive real has the argument zero ; every negative real the argument tt ; every positive pure imagi- nary the ai'gument ^ ; and every negative pure imaginary the argument — . "^ 2 . . • 21. If we add two imaginaries, the modulus of the sum is never greater than the sum of the moduli of the given imagi- naries. 12 INTEGRAL CALCITLIJS. [Art. 22. The sum of a + 5 V^ aud c +dV — 1 is a + c + (& + c?) V — 1. The moduhis of this sum is V(a + c)' + (& + ^0' ; t he sum o f the moduli of a +6 V^ and c +d V- 1 is V a^ + &' + Vc^ + d^. We wish to show that V(a + c)- + (6 + d)' -< Va^+F+ V"?T^; tlie sign -< meaning " equal to or less than." Now '\/{a + cy + {b + dy -< ^/'aF+¥-\- V c^ + d,\ if (a + c)2 + (& + cZ)2 -< cr + &' + 2V(fr + &") (c- + d-) +0^ + cZ^, that is, if ac + kZ -< Va^c^ + a^d^ + ft^c^ + bhl' ; or, squaring, if a' c' + 2 abed + 6^ d' -< a- (? + a' d- + &- c^ + ^ d' ; or, if -< {ad — be)-. This last result is necessarily true, as no real can have a square less than zero ; hence our proposition is established. 22. TJie modidns of the product of ttvo imaginaries is the product of the moduli of the given imaginaries., and the argument of the product is the sum of the arguments of the imaginaries. Let us multiply ?-i (cos (^1 + V — 1 . sin <^i) by ?-2(cos (^o + V — 1 . sin ^2) ; we get i\ >"2[cos (jbi cos ^2— sia <^i sin ^2+ V — l(sin <^i cos (^2+ cos <^i sin ^0)] , cos <^i cos ^2 •— sin ^1 sin ^2 = cos ( <^i + ^2) ? sin <^i cos ^2 + cos ^1 sin ^2 = sin (! + ^2) hj Trigonometrj' ; hence ri (cos <^i 4- V — 1 . sin (^1) r., (cos (^1 + V — 1 . sin (^2) = ri?'2 [cos (<^i 4- <^2) + V - 1 . sin ((^1 + (^o) ] , Chap. II.] IMAGIjSTAKIES. 13 and our result is in the tj-pical form, TiV^ being tlie modulus and •^1 + 4>2 the argument of the product. If each factor has the modulus unity, this theorem enables us to construct very easily the product of the imaginaries ; it also enables us to show that the interpretation of the operation V — 1 , suggested in Art. 18, is perfectly general. Let us operate on an}' imaginary subject, r(cos^4- V — 1. sin! + V — 1 . sin ^^ . To divide one imaginary by another, we have then to take the quotient obtained by dividing the modulus of the first by the modulus of the second as our required modulus, and the argu- ment of the first niinus the argument of the second as our new argument. 24. If w.e are dealing with the product of n equal factors, or, in other words, if we are raising r(cos<^ + V— l.sin^) to the 14 LNTEGEAIi CALCULUS. [Art. 25. nth power, n being a positive whole number, we shall have, by Art. 22, [r(cos ^ + V — 1 . sin ^) ]" = r"(cos?i + V — 1. sin?i cji) . [1] If r is unit^', we have merel}^ to multiply the argument by n, without changing the modulus ; so that in this case increasing the exponent by unit}' amounts to rotating the point represent- ing the imaginary through an angle equal to ^ without changing its distance from the origin. 25. Since extracting a root is the inverse of raising to a power, V[r(cos<^ + V^.sin<^)] = Trf cos^ + V^.sin^j ; [1] for, by Art. 24, -^•(eos-+V"^.sin-) ^ \ n ' nj — ?'(coS(^ + V — 1. sin<^). Example. Show that Art. 24 [1] holds even when n is negative or fractional. 26. As the modulus of every quantity, positive, negative, real, or imaginar}', is positive, it is alwa^'s possible to find the modulus of any required root ; and as this modulus must be real and positive, it ca7i never, in an}- given example, have more than one value. We know from algebra, however, that ever}' equa- tidn of tlie nth degree containing one unlinown has n roots, and that consequently ever}' number must have n nth roots. Our formula. Art. 25 [1], appears to give us but one nth root for any given quantity. It must then be incomplete. We have seen (Art. 20) that while tlie modulus of a given imaginary has but one value, its argument is indeterminate and may have any one of an infinite number of values whicli differ b}' multiples of 27r. If <^o is one of these values, the full form of Chap. II.] IMAGINAEIES. 15 the imaginary is not ?'(cos 4>o + V — 1. sin<^o) 5 ^.s we have hitherto written it, but is r [cos (^0 + 2 mir) -|- V — 1 . sin (<^o + 2 mir) ] , where m is zero or any whole number positive or negative. Since angles differing by multiples of 2 tt have the same trigo- nometric functions, it is easily seen that the introduction of the term 2 mir into the argument of an imaginar}^ will not modif}' any of our results except that of Art. 25, which becomes Vr [eos(<^o+ 2?ft7r) + V— 1- sin((^o+ 2?)i7r)] = ^/- Mo 2 7r\ , . Mo 27rY COS \-m — + v — 1 . sm \-m — [1] Giving VI the values 0, 1, 2, 3 , n — 1, w, n -\-l, success- ively", we get ^0 2 7r (j>Q 2tv 4>o 2Tr <^o , 5 1 5 1- ^ — 7 ho — 5 — + n—1 2- n n n n ^+2., ^ + - + 2., n n n as arguments of our 7ith root. Of these values the first n, that is, all except the last two, correspond to different points, and therefore to different roots ; the next to the last gives the same point as the first, and the last the same point as the second, and it is easil}' seen that if we go on increasing m we shall get no new points. The same thing is true of negative values of m. Hence we see that every quantity^ real or imaginary ^ has n distinct nth roots, all having the same modulus, but with argu- ments diflfering by multiples of ^^ . 27. Anj^ positive real differs from unity only by its modulus, and any negative real differs from —1 onl}' by its modulus. All the ?ith roots of any number or of its negative may be obtained 16 INTEGRAL CALCULUS. [Art. 27. hj multiplj'ing the nth roots of 1 or of — 1 by the real positive wth roots of the number. Let us consider some of the roots of 1 and of — 1 ; for ex- ample, the cube roots of 1 and of —1. The modulus of 1 is 1 and its aro-ument is 0. The modulus of each of the cube roots of 1 is 1, and their arguments are 0, — , and — ; that is, 0°, 120°, and 240°. The roots in question, then, are repre- sented by the points Pj, Pa, ^s, in the figure. Their values are l(cosO + V^. sinO), 1 (cos 120° + V^. sin 120°), and I(cos240° + V^.sin240°), or 1, -i+.^V^l, -i-^V^. The modulus of —1 is 1, and its argument is tt. The modulus of the cube roots of —1 is 1, and their arguments are ^, J + -^, !r + i5, that is, 60°, 180°, 300°. The roots in question, then, 3 3 are represented by the points Pi, Pg, Pg, in the figure. Their values are -^-\- 2 V — 1, — 1, -2 — 2 V — 1. Examples. (1) What are the square roots of 1 and — 1 ? the 4th roots ? the 5th roots ? the 6th roots ? (2) Find the cube roots of — 8 ; the 5th roots of 32. (3) Show that an imaginary' can have no real ?ith root ; that a positive real has two real ?ith roots if n is even, one if n is odd ; that a negative real has one real «th root if n is odd, none if n is even. Chap. II.] IMAGINAEIES. 17 28. Imaginaries having equal moduli, and arguments differing only in sign, are called conjugate imaginaries. r(cos<^ + V — l.siu^), and r[cos( — ) + V — l.sin( — ^)], or r(cos (^ — V— 1 . sin (^) are conjugate. They can be written x + y V — 1 and x — 7j\/—l, and we see that the points corresponding to them have the same abscissa, and ordinates which are equal with opposite signs. ExAjMPLES. (1) Prove that conjugate imaginaries have a real sum and a real product. (2) Prove, by considering in detail the substitution of a + 6 V— 1 and a — &V— 1 in turn for x in an}^ algebraic poly- nomial in X with real coefficients, that if any algebraic equation with real coefficients has an imaginary root the conjugate of that root is also a root of the equation. (3) Prove that if in an}^ fraction where the numerator and denominator are rational algebraic polynomials in x, we substi- tute a + 6V— 1 and a — &V — 1 in turn for x, the results are conjugate. Transcendental Functions of Imaginaries. 29. We have adopted a definition of an imaginary and laid down rules to govern its use, that enable us to deal with it, in all expressions involving only algebraic operations, precisely as if it were a quantity. If we are going further, and are to sub- ject it to transcendental operations, we must carefully define each function that we are going to use, and establish the rules which the function must obey. The principal transcendental functions are e^, logic, and sinx, and we wish to define and stud}' these when x is replaced by an imaginary variable z. As our conception and treatment of imaginaries have been entirely algebraic, we naturall}' wish to define our transcendental 18 INTEGRAL CALCULUS. [Art. 30. functions b}' the aid of algebraic functions ; and since we know that the transcendental functions of a real variable can be ex- pressed in terms of algebraic functions onlj' by the aid of infinite series, we are led to use such series in defining transcendental functions of an imaginary variable ; but we must first establish a proposition concerning the convergency of a series containing imaginar}' terms. 30. If the moduli of the terms of a series containing imaginary terms form a convergent series, the given series is convergent. Let Uq -{- III -\- ih -^ + w» + be a series containing imagi- nar}' terms. Let ^(o = i2o(cos*o+ V — 1. sincEJo), ti^ = i?i (cos $i + V — 1 . sin ^j), &c., and suppose that the series i?o + -Ki +-R2 + + -S« + is convergent ; then will the series Uq-}- ri+i?2sin$2-l • + i2„sin$„-f ). (2) (1) can be separated into two parts, the first made up only of positive terms, tli€ second only of negative terms, and can therefore be regarded as the difference between two series, each consisting of positive terms. Each term in either series will be a term of the modulus series ^0+^1 + ^2 + multiplied by a quantit}' less than one, and the sum of n terms of each series will therefore approach a definite limit, as n increases indefi- nitely. The series (1), then, which is the abscissa of the point representing the given imaginary series, has a finite sum. Chap. II.] IMAGINAEIES. 19 In the same way it ma}'- be shown that the coefficient of V— 1 in (2) has a finite sum, and this is the ordinate of the point representing the given series. The sum of n terms of the given series, tlien, approaches a definite limit as n is increased indefi- nitely, and the series is convergent. 31. We have seen (I. Art. 133 [2]) that e^ = l+^-f-^ + ^-)-^+ [1] 12!3!4! when X is real, and that this series is convergent for all values of x. Let us define e% where z = x-{-y\/ — l, by the series 12! 3! 4! ■- -^ This series is convergent, for if z = r(cos ^ + V— 1 . sin <^) the series i + !:+i!+il+i!.+ 1 2! 3! 4! made up of the moduli of the terms of [2] is convergent hj I. Art. 133, and therefore the value we have chosen for e" is a determinate finite one. Write a; + ?/V— 1 for 2, and we get c'+v^/=i^l I a^+yV-l ^ (a;+yV3i)2 ^ (a;_^yV~i)3 ^ The terms of this series can be expanded by the Binomial Theorem. Consider all the resulting terms containing any given power of X, say x^ ; we have pr^ 1 ^ 2! ^ 3! ^ ^ ;:! ^""r or, separating the real terms and the imaginary terms, ^(i_i! + ^_/ + >, 2^1 2!4! 6! ^ pi ^^ 3! 5! 7! ^' 20 INTEGRAL CALCULUS. [Art. 32. or — (cosy + V^. sin?/), b}' I. Art. 134. p ! Giving p all values from 1 to co we get gx+y./^ = (cos ?/ + V^. siny) (i+Y + fr + fr + fT'^ ^ = e^ (cosy + V—1. sin?/), [4] which, b}' the way, is in one of our tj-pical imaginary forms. If a;=0, in [4], we get e^^-i = cos?/ + V^.siny, which suggests a new way of writing our typical imaginary; namely, * r (cos ^ + V — 1 . sin ^) = re*^^-\ 32. We have seen that let us see if all imaginary powers of e obey the laiv of indices; that is, if the equation g«g«^g« + « ["ij is universally true. Let ?( = a;i + ?/iV— 1 and 'y = 3^2 + 2/2^—1, then e"= e^i + 2/1 -^^ = e«i (cos?/i + V— 1 • sin?/i) , 6"= ea:o + 2/oN/^= ea^^^-cosya + V— 1. siuyg) ? g« g« _ ga;i ga;^ ["cog (^/^ -f 3/2) + V — 1 • sin (?/i + y,) ] = e^i+ ^'2 [cos (yi + Vi) + V^ . sin (yi + 2/2) ] _ ga;i + a;2 + (2/1 + 2/2) V^ __ gM + » and the fundamental property of exponential functions holds for imaginaries as well as for reals. > Example. Prove that a'^a" = ^4" + " when u and v are imaginary. Chap. II.] IMAGIISTAKIES. 21 LogaritJimic Functions. 33. As a logarithm is the inverse of an exponential, we ought to be able to obtain the logarithm of an imaginar}- from the formula for e'=+^"'^^ We see readily that z = r (cos<^ + V^. sin<^) = eiogr+0^ whence logz = logr + ^ V — 1 ; or, more strictly, since z = r[cos (<^o + 2n7r) + V — l.sin(o+ 2n7r)], •log2 = logr+(riori, for g27rv/^i^gQg27r+V^.sin27r= 1, by Art. 31. Hence, adding 2 ir V— 1 to the logarithm of an}^ quantity will have the effect of multiptying the quantity by 1 , and therefore win not change its value. , Example. Show that if an expression is imaginary, all its logarithms are imaginary ; if it is real and positive, one logarithm is real and the rest imaginary ; if it is real and negative, all are imaginary. 22 INTEGRAL CALCULUS. [Akt. 34. Trigonometric Functions. 34. If z is real, sin. = . -- + --_ + [1] eos. = l-|l + |l-^+ [2] byl. Art.134. ^- ^- '^ ' If 2 = r(cos^+ V— l.sinc^), the series of the moduli, n\0 ntO rt»7 ^" + FT + ^ + ^+ , 3 ! ! 7 ! 1 + if + j1 4. i! _f- 2 ! 4! 6! ' are easily seen to be convergent ; therefore if z is imaginary, the series [1] and [2] are convergent. We shall take them as defi- nitions of the sine and cosine of an imaginary. Example. From the formulas of Art. 31, and from Art. 34 [1] and [2], show that gzV^ _ COS2 _|_ V — 1. sin2, and e'^"^"^ = coss; — V — 1. sin^, for all values of z. 35. From the relations g^v-i _ gQg J, _|_ y_][^ sins;, we get C0S2 = — , [1] sm2 = — , 12 I for all values of 0. Chap. II.] Let cos(cc+?/V— 1) = IMAGINAEIES. z — x + y'\l —1. g^v/-l_y^g-x^^l + 2 23 _ (cosx+V— l.siaa;)e~^+(cosa;— V— l.sina;)e^ ~ 2 ' by Art. 34, Ex., = COS a; — ■ — V — 1 . siiiic 2 In the same wa}^ it ma}^ be shown that [3] . , I — 7^ (cosa;+ V — 1. sina;)e"^— (cosa; — V— l.sina;)e^ l(.x+?/V-1)=^ ~ '- — ^ — 2V-I + V-1. cos X ■ [4] If z is real in [1] and [2] , we have cos X = + e- 2 „xV-l „-lV^l V^. lfz = y V — 1, and is a pure imaginary, cosy 1 = 2 sin ?/ V — 1 = 2 V^; [5] [6] whence we see that the cosine of a pure imaginary is real, while its sine is imaginary. By the aid of [5] and [6] , [3] and [4] can be written : cos (x-\-yy/ — 1) = cosoJcosyV— 1 — sinxsin?/ V— 1, [7] sin (x + 2/ V— 1) = siniccos?/ V— 1 + cosccsin^/V — 1. [8] 24 INTEGEAL CALCULUS. [Art. 36. Examples. (1) From [1] and [2] show that sin^2:+ cos^2;= 1. (2) Prove that cos (?i -{-v)= COS M cos 'y — shaitsmv, sin (u -\-v) = sinttcosi; + cosMsmi>, where u and v are imaginary. The relations to be proved in examples (1) and (2) are the fundamental formulas of Trigonometry', and they enable us to use trigonometric functions of imaginaries precisely as we use trigonometric functions of reals. Differentiation of Functions of Imaginaries. 36. A function of an imaginary variable, is, strictly speaking, a function of two independent variables, X and y ; for we can change z by changing either x or ?/, or both X and y. Its differential will usually contain dx and dy, and not necessarily dz ; and if we divide its differential by dz to get its dy derivative with respect to 2, the result will generallj- contain —■, OjOC which will be wholl}'^ indeterminate, since a; and y are entirely independent in the expression x + y V — 1. It may happen, however, in the case of some simple functions, that dz will appear as a factor in the differential of the function, which in that case will have a single derivative. 37. In differentiating^ the V — 1 may be treated like a con- stant; for the operation of finding the differential of a function is an algebraic operation, and in all algebraic operations V — 1 obej^s the same laws as an^^ constant. CH-iP. II.] EMAGIiSrAEIES. 25 Example. Prove that d(xV^) = 2x V^ . dx ; and that dV— 1. sma;= 'sj — l.cosx.dx We have, by the aid of this principle, if z = x-\-y\r—i, dz= dcc + V — 1. dy; [1] if z = r (cos<^ + V — 1. sine/)), dz = d9'(cos ) (cos <^ + V^ . sin cjy). [2] 38. Let us now consider the differentiation of 2"', e^, logs, sins, and cos 2;. Let z = r (cos <^ + V— 1 . sin <^) , then gm _ ,''»(cos?n^ + V^. sinm^) , hj Axt. 24 [1] ; dz"" = jjir™"^ dr (cos 77i<^ + V — 1 . sin ??i ( — sin m <^ + V — l.cosmc/)), d^'" = 7/ir™-^ [cos (7)1—1) ] (cosej!) + V — 1. sin (j>)dr + m?''"[cos (7?i— 1) <^ + V — l.sin(m — 1)0] (coscfi + V^. sin<^) ^f^.dcji, dz^ = mr""'^ [cos (m — 1 ) c/) -f- V — 1 . sin (m — 1 ) (^] (dr _j- ?-V — 1 .dcf>) (cos + V— 1. sin<^), dz"' = mz'^-hlz, [1] byArt. 37[2], ^ = mz"'-\ [2] and a power of an imaginarj^ variable has a single derivative. Example. Show that [1] and [2] hold for all powers of z. 26 ESTTEGEAL CALCULUS. [Art. 39. 39. If 2=a; + ?/V^, e'^ = e'^(cos2/+V — l.sin?/), by Art. 31 [1], de' = e'dx ( cos ?/ -f V^ . svay) -\- e" {— sin y -\-'\/—l.cosy)dy, de^ = e==(cos2/+V — 1. sin?/) (dx+V— 1. dy), de^=e''dz, [1] ^^' = e^ ' [2] d^ Show that Example. da'^ = a^ los: a. dz. 40. If « = ^(cos<^ + V-l.sin(^), log2: = log?' + ^ V — 1, by Art. 33, d\ogz = 1- V — 1 . d<^ = ■ ^, g^l^^^ ^ (r7r+rV-l.r?.^)(cosj>+V-l.sin<^) r(cos(^ + V — 1. sin^) d loaf 2 = dz dlogz _ 1 dz z [1] [2] 41. sins:: dsins; = 2V-1 by Art. 35 [2], ^.^i_^^_.^i 2 V-1 g^«Ari_|_g_^V_i V-1.CZ2 dz, dsin2 = cosz.dz. by Art. 35 [1], [1] Chap. II.] niAGINAEIES. 27 cosz = , dcos« = ^ ^ V-l.cfe = —=z cfe, 2 2 V^ dcosz = — sin2!.cfe. [2] 42. "We see, then, that we get the same formulas for the dif- ferentiation of simple functions of imaginaries as for the dif- ferentiation of the corresponding functions of reals. It follows that our formulas for du'ect integration (I. Art. 74) hold when x is imaginar}^. Hyperbolic Functions. 43. We have (Art. 35 [5] and [6]) cos a; V^ = ,^' + ^"' 2 ' and sina? V^ ^ e" - ^"V ^, 2 where a; is real. — r — is called the hj^perbolic cosine of x, and is written Ch x ; and — - — is called the hj^perbolic sine id of x^ and is written Sha; : Sha; =. ^^ ~ ^~' = - V^. sina; V"^, [1] Cha; = ^' + ^~' = cos a; V^. [2] The hj^perbolic tangent is defined as the ratio of Sh to Ch ; and the hjiDerbolic cotangent, secant, and cosecant are the re- ciprocals of the Th, Ch, and Sh respectively. These functions, which are real when x is real, resemble in their properties the ordiuar}^ trigonometric functions. 28 INTEGRAL CALCULUS. 44. For example, for and Ch2a:-Sh2a;=l; gSx _|_ 2 ^ Q-2Z Ch^x Sh2£c = e2x_ 2 _|_ e- [Art. 44. [1] Examples. (1) Prove that 1 -Tli^'c= Sch^'c. (2) Prove that 1 - Cth2aj= Csch^o;. (3) Prove that Sh(iB + ?/)= ShicCh2/ + Cha^Sh?/. (4) Prove that Ch(a; + ?/) = Ch.'cCh?/ + ShxShy. 45. dSha; = d- e' -\-e' dx. dShx = Chx.dx. Examples. (1) Prove dChx = Shx.dx. dThx= Sch^x.dx. dCtha;= —Csch^x.dx. ,. dSehcc= —SehxThx.dx. dCschx = — Cscha;Ctha.\da;. 46. We can deal with anti-h}^erbolic functions just as with anti-trigonometric functions. To find dSh-^a;. Let then u — Sh'^o;, x= Sh^t, dx— Chw.cZw, Chap, n.] LMAGESTAEIES. 29 du= , Chu Chu= Vl + Sh^w, by Art. 4 [1] , Examples. dCh-^x = dTh-^x = Prove the formulas dx \ CI Sell ''^x = — dCsch.~^x=— ^/x''-l dx dx x Vl— 0^ dx X Var + 1 47. The anti-hj-perbolic functions are easily expressed as logarithms. Let ic=Sh~'^x, then a; = Sh tt = e" — e' 2x = e" , e"-x= ± VlTa?, e" = « ± VIT^ ; 30 ESTTEGBAL CALCULUS. [Akt. 48. as e" is necessarily positive, we may reject the negative value in the second member as impossible, and we have u = log(.i; + Vl + it--) , or Sh-iic = log(x4- Vr+^). [1] EXAJEPLES. Prove the formulas Ch-^a; = log(.T + Va;- — 1) . Th-^aj = ilog l+x 1 — X Csch-'a;=Iog('l + Ji+A 48. The principal advantage ai'isiug from the use of h}^er- bolic functions is that they bring to light some curious analogies between the integrals of certain irrational functions. From I. Art. 71 we obtain the formulas for direct integration. s dx . _i rt-t = sm ^x. [1] V 1 - ar '^"^ =tan-ia;. [2] 1 + it^ dx X Va;^— 1 From Art. 46 we obtain the allied formulas dx sec"^a;. [3] Vl+a.-^ dx = Sh-^x = log{x + VI+^). W = Ch-ia; = log(a; + V^^^^). E^] Chap. 11.] EVIAGINAEIES. r dx J l-x^ r dx -=Sch-ia) =looYi4-^ x^l—x^ 31 [6] [7] - f^^ = Csch-^^ = log fi + aE+iY [8] 32 INTEGRAL CALCULUS. [Art. 49. CHAPTER III. GENEKAL METHODS OF INTEGEATING. 49. We have defined the integral of any function of a single variable as the function which has the given function for its derivative (I. Art. 53) ; we have defined a definite integral as the limit of the sum of a set of differentials ; and we have shown that a definite integral is the difference betioeen two values of an ordinary integrcd (I. Art. 183) . Now that we have adopted the diflTerential notation in place of the derivative notation, it is better to regard an integral as the inverse of a differential instead of as the inverse of a derivative. Hence the integral of fx.dx will be the function whose differ- ential is fx.dx; and we shall indicate it by i fx.dx. In our old notation we should have indicated precisely the same function by I fx ; for if the derivative of a function is fx we know that its differential is fx.dx. 50. If /c is an}' function whatever of a;, fx.dx has an integral. For if we construct the curve whose equation is y=fx, we know that the area included by the curve, the axis of X, any fixed ordinate, and the ordinate corresponding to the variable x, has for its difl'erential ydx., or, in other words, fx.dx (I. Art. 51). Such an area always exists, and it is a determinate function of x, except that, as the position of the initial ordinate is wholl}' arbi- trary, the expression for the area will contain an arbitrary con- stant. Thus, if Fx is the area in question for some one position .of the initial ordinate, we shall have Cfx.dx = Fx-\-G, where C is an arbitrary constant. Chap. III.] GENERAL METHODS OF INTEGEATING. 33 Moreover, Fa; + C is a complete expression for j fx.dx ; for if two functions of x have the same differential, they have the same derivative with respect to x, and therefore they change at the same rate when x changes (I. Art. 38) ; they can differ, then, at any instant only by the difference between their initial values, which is some constant. Hence we see that every expression of the form fx.dx has an integral^ and, except for the presence of an arbitrary constant, but one integral. 51. We have shown in I. Art. 183 that a definite integral is the difference between two values of an ordinaiy integral, and therefore contains no constant. Thus, if Fx-\-G is the integral of fx.dx, Jfx.dx = Fb — Fa. a In the same way we shall have Cfz.dz = Fb-Fa; and we see that a definite integral is a function of the values between tohich the sum is taken and not of the variable with respect to which we integrate. Since ffx.dx^Fa — Fb, I fx.dx = — I fx.dx. h Ja Example. fx.dx + I fx.dx = I fx.dx. 52. In what we have said concerning definite integrals we have tacitly assumed that the integral is a continuous function between the values between which the sum in question is taken. If it is not, we cannot regard the whole increment of Fx as equal 34 INTEGRAL CALCULUS. [Art. 53. r to the limit of the sum of the partial infinitesimal increments, and the reasoning of I. Art. 183 ceases to be valid. Take, for example, ) — -. and apparently j-=j.-c?. = _ = --, by I. Art. 55 (7); But j — ;- ought to be the area between the curve ?/ = — , the axis of X, and the ordinates corresponding to cc = 1 and ic = — 1 , which evidently is not — 2 ; and we see that the function — is discon- x^ tinuous between the values x= —I and x = \. The area in question which the definite integral should represent is .J- easil}' seen to be infinite, for -1 J-\ ar e J e x^ €. and each of these expressions increases without limit as e ap- proaches zero, 53. Since a definite integral is the difference between two values of an indefinite integral, what we have to find first in any problem is the indefinite integral. This ma}' be found by in- spection if the function to be integrated comes under any of the forms we have already obtained b}' differentiation, and we are then said to integrate directl}". Direct integration has been illus- trated, and the most important of the forms which can be in- tegTated directly have been given in I. Chapter V. For the sake of convenience we rewrite these forms, using the diflferential notation, and adding one or two new forms from our sections on hyperbolic functions. Chap. III.] GENERAL METHODS OE INTEGEATIKG. 35 /^,n + 1 x''dx = - . 71 + 1 /dx , — = logic. X /a'dx=z-^-. loga I e'dx^ e". I smx.dx= —cos a;. I cosx.dx = sin a;. I tana.\cZa;= — logcosa;. I ctua;.da;= logsino;. I = sm ^a;. c/ Vl-ic^ I , = Sh-^a; = log(a; + Vl + x^) . *^ Vl + a^ /- . = Qh.-'^x = log (a; + V.«^ — 1) . Va.-^ — 1 /r^-Th-..=iiogi±|. /- da; , = sec~^a;. ■■"s/a?— 1 /-^ = - sch-.. = - log fi + J^:^). ^ x^ll~x^ \x \a? J r do^ ^ _ ^^^^_,^ ^ _ A JT^Y dx ^2x — ^ C dx , I . = vers~-^a;. 36 INTEGEAL CALCULUS. [Art. 54. 54. We took up in I. Chap. V. the principal devices used in preparing a function for integration when it cannot be integrated directly. The first of these methods, that of integration by substitution, is simplified by the use of the difl'erential notation, because the formula for change of variable (I. Art. 75 [1]), i u= I uDyX becoming | udx— i u — cly, reduces to an identity'' and is no longer needed, and all that is requked is a simple substitution. rdx , (a) For example, let us find I — VI + logaj. Let 1+ logic = 2;, then — = a2 ; X aud J -^ Vl + logaj =J zldz = %zl = f (1 + logo;)!. (&) Required j - dx Let e* = y, then ^dx — dy, dx e'dx dy gx _|_ g-x g2x ^ 1 y2_^-^ and f-T^^x = Tt^^ = tan-i^/ = tan.-V J e^ + e-" J l+y^ "^ (c) Required | SQCx.dx. Let then seca; = 1 coscc COSflJ cos ^03 z = sin a; > dz- = cos a;. dx. cos' '■x = \- -z\ Chap. III.] GENEEAL METHODS OF INTEGRATING. 37 i— = I :; i = ilog -^— , by Art. 53, cos ''a; J 1 — z^ 1 — z secic.aa; = flog 1 — sina; Examples. Prove that (1) Ccscx.dx = ^log ~ — ^^^ = logtan - J 1 + cos a) 2 x'dx 1 T a; Vl — a;^ := — ^cos~^a; Suggestion : Let x = cos 2;. 55. The formula for integration by parts (I. Art. 79 [1]) becomes I udv = uv — I vdu, - [1] when we use the differential notation. It is used as in I. Chap. V. (a) For example, let us find I a;" log a;, da;. Let w = loga;, and dv = x''dx\ then du = — , x and v = «;• n + l n + l a;"loga;.aa; = loga; — I dx = f los-o: 1. ^ n + \ ^ J n+l n + l\ "^ n-{-l) (6) Required | a;sin~^a;.dx. Let if=sin~-'a;, and dv = xdx ; then du= ^^ . INTEGRAL CALCULUS. [Art. 55. and '" ~ ^' I xs,ixr^x.ax = — sin a; — -I- 1 — , X sin"-^ x.dx=^ — sin~^ 'x-\-\ (cos"^ x + x V 1 — oj^) . (c) Eequired i — x^dx Let w = a;e'', dec and cZv then du = (aje'' + e"") (^a; = e^( 1 + a?) c?aj, 1 and 1+aj {\-VxY 1+a; J 1 + a; 1 Examples. (1) I — — = sm^ — i-^^. -^ Vl-3x-a^ Vl3 (2) I a;tan"-^a;.dx = — i^— tan~^a; — -^-a;. ^ ^ J {1-xy^ ~ 1-x 2{i-xy' (4) I — = — V2 ga; — a^ + a vers~-^-. »^ V2 aa; — 3/*^ (^ (5) rV2 aa; - a:^. da; = ^ V2 ax - ar^ + ^' sin'^ '^^^. ^ 2 2 a Suggestion : Throw 2 ax — a;^ into the form a^ — (x — a)^. (6) I -^ dx = log(x -j- sinx) . J X -{- sinx Chap. III.] GENERAL METHODS OF mTEGEATING. 39 (/) I — ■ c(a;=£ctan — */ 1 + cosic 2 Suggestion : Introduce — in place of x. J xijogxy^ ~ ~ {n~l) (logcc)"-^" (9) nog(logx) ^^ ^ j^g^. [log(loga;) - 1] . (10) ( ^ = 2;tan2; + logcos2;, wliere2 = sin~^cc. J ( 1 — or) a (11) r^^J^ = 1 logtan f^ + 'L). J sinaj+cosa; Y2 \2 8/ 40 INTEGKAL CALCULUS. [Art. 56. CHAPTER IV. EATIOFAL FRACTIONS. 56. We shall now attempt to consider sj'stematically the methods of integrating various functions ; and to this end we shall begin with rational algebraic expressions. Any rational algebraic polynomial can be integrated immediately hy the aid of the formula X" s- x^dx = n + 1 Take next a rational fraction^ that is, a fraction whose nu- merator and denominator are rational algebraic polj'nomials. A rational fraction is proper if its numerator is of lower degree than its denominator ; improper if the degree of the numerator is equal to or greater than the degree of the denominator. Since an improper fraction can alwaj^s be reduced to a pol^'nomial plus a proper fraction, b}' actually' dividing the numerator b}' the denominator, we need only consider the treatment of proper fractious. 57. Every proper rational fraction can be reduced to the siim of a set of simpler' fractions each of which has. a constant for a numerator and some poiver of a binomial for its denominator; that is, a set of fractions any one of which is of the form ■ fx Let oui' given fraction be -^^ • ^ Fx (x— a)' If a, &, c, &c., are the roots of the equation, i^a; = 0, (1) we have, from the Theor}' of Equations, Fx =A{x- a) (x - 6) {x - c) (2) Chap. IV.] RATIONAL FRACTIONS. 41 The equation (1) may have some equal roots, and then some of the factors in (2) will be repeated. Suppose a occurs p times as a root of (1), b occurs q times, c occurs r times, &c., then Fx = A{x-ay{x-by{x-cy (3) Call A{x — by{x — cy-- = (fiX', then Fx = (x — ay^tx, fa fa J. ~ fx ;— d>X -. — Ax and ^ = ^ = ^f^+ -"" Fx {x — ay cfix {x — ay (f)X {x — ay ^x fa . fa ^ -r~ fx ; — (bX 4>a •' <^a ^ fa {x — ay {x — ay<^x f^~T^^^ —, is a new proper fraction, but it can be reduced {x — ay(l>x ^ ^ to a simpler form by dividing numerator and denominator bj'^ a; — a, which is an exact divisor of the numerator because a is a root of the equation fx—J—<^x= 0. If we represent b}' fiX the quotient arising from the division of fx — ^ (fiX by X — a, we shall have fa fx ^ (jyq fiX Fx {x — ay {x — a)"-^ cjiX ' •fry* ' where ^ = — is a proper fraction, and may be treated {x — a)^~^ (jiX precisel}^ as we have treated the original fraction. fa Hence i^ = il + __.^^ . {x — ay-'^x {x — ay-'^ {x — ay-^(l)X By continuing this process we shall get fa fa_ fa f^.^a fx ^ (i>a f^a 4>a . a f^x Fx {x — ay{x—ay-^{x — ay-'^ .x — ax' 42 INTEGEAL CALCULUS. [Art. 58. In the same wa}' -^ can be broken up into a set of fractions cf)X having (x — b)^, (a; — 5)«~S &c., for denominators, plus a frac- tion which can be broken up into fractions having (x—cy, (^x — cY'^ , &c., for denominators; and we shall have, in the end, Fx {x — ay {x- ay-^ '^ x-a {x-by ^' ■ ^" ■ •• + -£; [1] {x — by~^ x — b where K is the quotient obtained when we divide out the last factor of the denominator, and is consequent!}* a constant. More than this, K must be zero, for as (1) is ideuticall}' true, it must fx be true when x=co ; but when x= cc I — becomes zero, be- Fx cause its denominator is of higher degree than its numerator, and each of the fractions in the second member also becomes zero; whence K=0. 58. Since we now know the form into which any given rational fraction can be thrown, we can determine the numerators by the aid of known properties of an identical equation. Let it be required to break up -^-^ — into simpler iX — ^j" \X "i~ i- ) fractions. By Art. 57, 3a; -1 A , B , O {x-iy{x-^l) (a;-l)- x-1 aj + 1' and we wish to determine A, B, and C. Clearing of fractions, we have 3x-l=^A(x+l) + B(x-l) {x+l) + C(x-iy. (1) As this equation is identically true, the coefficients of like powers of x in the two members must be equal ; and we have 5 + C=0, A-2C=3, A-B + C=-l; Chap. IY.] EATIONAL FEACTIONS. 43 whence we find -4=1, C=-l; and __3^^1 =_^^+J L.. (2) The labor of determining the required constants can often be lessened by simple algebraic devices. For example ; since the identical equation we start with is true for all values of x, we have a right to substitute for x values that will make terms of the equation disappear. Take equa- tion [1] : 3x-l = A{x+l) + B{x+l){x-l) + C{x-iy. [1] Letx' = l, 2 = 2 A, A=l, then 2x-2 = B (x + 1) (x -l)-\-C {x-iy ; divide by oj-l, 2 = B (^x + l) + C (x-l). Leta;=l, 2 = 25, 5=1, then —x + l=C{x-l), C=-l. Examples. (1) Show that when we equate the coeflficients of the same powers of x on the two sides of our identical equation, we shall always have equations enough to determine all our required numerators. (2) Break up — -^^ into simpler fractions. ^ ^ ^ {x-3y{x-j-l) ^ 59. The partial fractions corresponding to any given factor of the denominator can be determined directly. 44 INTEGRAL CALCULUS. [Akt. 59. Let us suppose that the factor in question is of the first degree and occurs but once ; represent it b}^ x — a. =^=-^ + ^ (1) Fx X — a (f>x'' ^ ^ by Art. 57, where Fx x^ ^ where Fx = (x—aY x. fx Multiply (4) by {x — a)", and represent (a; — a)" ^ by f^x. ^x=A^ + A^ix — a) +As{x — ay + + A^{x — a)""^ ^x 4.!A^(a;_a)". Chap. IV.] EATIONAL FRACTIONS. 45 Differentiate successively both members of this identity, and put x=a after differentiation, and we get Ai = $a, 2! " Ai = —^>"a, 3! A= (n-l) ■ $("-^>a. Although these results form a complete solution of the prob- lem, and one exceedingly neat in theoryf the labor of getthig the successive derivatives of ^x is so great that it is usually easier in practice to use the methods of Art. 58 when we have to deal with factors of higher degree than the first. So far as the fractions corresponding to factors of the first degi'ee are con- cerned, the method of this article can be profitably combined with that of Art. 58. 60. As an example where the method of the last article applies well, consider Sx-1 = ^+_A_+ ^ x(x-2){x + l) X x-2 ic + l' 3a;-l A = C = Scc-l _{x-2){x-^l) ^1 1 = 2 '_3£-l_"| ^5 x{x-\-l)_^^2 6 ' 3a;-l " _x(x — 2) = i 1,5 1 4 3' a;(a; — 2)(a;+l) 2 x 6 x-2 3 a^ + l [1] 46 INTEGRAL CALCULUS. [Art. 61. Again, consider 1 1 _ A B l + x^ (a; + V-l) (a;-V-l) iC+V-1 x-^T- -1 A — 1 1 v^ _X—'\J — \_ , = _vn 2V-1 2 ' B — 1 1 V-i Lcf + V-i_ ^=^-1 2V-1 2 1 V-i 1 V-i 1 r l + aj2 :+V-l. :-V-l [2] 61. Let us now consider a more difficult example, where it is worth while to combine our methods. To break up ~ ^ {x-iy{x^+l) a;3 + 1 = (x + 1) (a^ - a; + 1) = (a:+l)(x-i-|V^)(x-i + iV^3), a^ + 1 _ ^'' + 1 ^1 {x-iy{x^+l) {x-iy{x + l){ar-x+\) {x-iy I Ao . A^ . Aj B . C ^ (^x-iy {x-iy'^x-i'^x+i'^^_^_x^z:s -i+iV,-3 (1) B = C = D = af+1 {x-iy{x''-x + i) g;^ + 1 _{x+l){x^ — x+l)_ x^+1 _ 1 :=-x"24' = 1, i{x-iy(x+l){x-l-^i^-3)j x^+1 :=i+i,sAr3 _{x - ly (x + l){x-i- i V-3)>|-| v-3 Chap. IV.] BATIONAL FEACTIONS. 47 _1 _1 3 3 1 (2a;-l) Substitute these values and clear (1) effractions. 24.{x^-{-l) = 24.{x-{-l){x'-x+l)+24.A.Xx-l){x+l){x'-x-\-l) + 24.As(x-ly(x■j-l)(af-x+l) + 24.A^(x-ly(x + l) (a^-.T+l) + (x-iyix'-x+l) -8(2x--l) (a;- 1)^^; +1) ; 15x'^-olaf+A5x*+6x^-51x^+i5x-9=24:A2{x-l){x4-l) (a;2 _ ic + 1) + 24.43 (x - 1)^ {x + 1) (a^ - a; + 1) + 24^^ (x-iy(x + l) (x^-^x-1). The second member of this equation is divisible by (^a;-l)(x+l)(x'-x-{-l), therefore the first member must be divisible b}' the same quantit3\ Dividing, we have 15a;2- 3607 + 9 = 24^2 + 24^3(0; -l)+24^4(a;-l)2. Leta;=l, — 12 = 24 Jla, 2 2' and we get 15^•2-36.^' + 21 = 24^3(cc-l) + 24^4(x-l)^ Divide b}^ cc — 1 ; 15a; -21 = 24^3 + 24^4(a; - 1) . Leta.'=l, —6 = 24^3, A --^ ^3 - - p 15a;-15 = 24^4(a;-l). Divide by a; — 1 ; 15 = 24 J.4, 48 INTEGRAL CALCULUS. [Art. 62. Hence :^ + l _ 1 1 1 1 1 .51 (x-iy{x' + i) {x-iy 2\x-iy a' {x-iy s'x-i + 1.J: i. ' _ -l 1 _ . (2) 24 x + 1 3 rc-i-iV-S 3 a;-i + iV-3 62. Having shown that any rational fraction can be reduced to a sum of fractions which alwa^^s come under one of the two A A forms and , it remains to show that these forms (x~aY x—a can be integrated. To find f ^^^ J {x — a) let z = x — a, then dz = dx., and r Adx ^j^Cdz^ \ A ^ 1 A PJ-, To find f ^'^'^ , J X — a let z = x— a. then dz = dx, and r-^^=^r— = ^log2 = ^log(a;-a). [2] Turning back to Art. 58 (2) , we find / (3a; — l)da; _ r dx r dx _ C dx _ _ 1 {x-\y{x + l)~J {x-iy J x-l J x+\~ x-1 + log(a;-l)-log(a; + l) = -- — - + log^^. Turning to Art. 60 (1), we have / {^x—\)dx ^ jL Cdx 5 r dx _ 4 r dx x{x-2){x-l)~^J X ^J x-2 V a; + l = iloga; + f log(a; - 2) - Alog(x + 1). Chap. IV.] RATIONAL FRACTIONS. 49 63. If imaginary values come in when we break up our given fraction, they will disappear if we combine oUr results properly after integrating. We know (Art. 28, Ex. 2) that if the denominator of our given fraction contains an imaginary factor, (x — a — b V— 1)", it will also contain the conjugate of that factor, namel}', (» — a + 6 V— 1)". Moreover, since b^- Art. 59 the numerator of the partial fraction corresponding to (a; — a — 6 V — 1)" will be the same rational algebraic function of « + 6 V — 1 that the nu- merator of the partial fraction corresponding to {x — a-{-b V— 1 )" is of a — 6 V — 1, these two numerators must be conjugate imagi- naries by Art. 28, Ex. 3. Hence, for every fraction of the form ^^t I I — — we shall have a second of the form (a; — a — 5 V — 1)" (cc — a+&V — 1)" J A + bV^ ^,^^ 1 (A + B-J^l) f-, (.x--a-6.V-l)" 0^-1) (ic-a-S V— 1)"-^ by Art. 62 [1], A-B^~l ^i^^_ 1 {A-B-J~\) _ (x — a + &V — 1)" O'' — 1) (a; — a + 6V — l) Let (a;-c6 + 6V^)"-' = X+rV'^, X and Y being real functions of x ; then (a;-a-&V^)"-' = X-rV^. A + B^^l , ,r A-Byf^l r A + B^-^ ax+C-A ^ (x — a—h^—iy -J (x — dx (a;-a-6V-l)" -^ (x - a + & V-1)' ^ 1 {A + B V^) 1 (A-B V^) (n-l)' X-r^^l (n-1) x+fV^ = 1 (2AX+2BY) ~ (n-l)' (.-c- - 2 ax + a' + &')""'' a result which is free from imaginaries. [1] 50 INTEGRAL CALCULUS. [Art. 63. If n=l, we have the pair of fractions, ■ 1:^:; and x—a — hyJ — 1 x—a+h'\/ — l r A + B-sJ-l c^a; = (A + ^ V^)log(a;-a-6V3T). •^ cc — a — 6 V— 1 by Art. 62 [2], r_A--B^{^__ ^j^ ^ (^ _ 5 VHI) iog(ic - a + 6 V^^) ; ^ x—a+b V— 1 log(a; - a - 6 V^) = ^log [(« - ay + 6^] _ V^. tan"^ . ^ a;— a ic — a log(aj-a + &V^) = ilog[(cc-a)2 + &2] + V-l.tan-^—^ a; — Hence f-^+^V^ ,,^ + r^-i?V^ ^, *^ x — a — b\/—l ^ x — a -\-by—l = ^log[(iB-a)2 + &2] + 25tan-i-^, [2] » — a which is real. The form of [2] can be modified by adding a constant. "L 4. tan-1 -^ - - 4- etn-^ ^"^^ - !!: _ etn-^ ^^-^ ' - tan-i ^'' ~^ 2 x-a 2 & 2 6 6 Hence Alog[(x-ay + b''] +2Bt2in-^^^^:^ [3] differs from [2] by the constant jBtt, and therefore is a true value of r_^+WHL0 and j:>>0, a repeated use of [5] will reduce p to zero, and we shall have to find merely the I x'"-~^dx. Chap. V.] KEDUCTIOlsr FOEMULAS. 55 If m<0 and p>0, [3] will enable us to raise m to 0, and then [5] will enable us to lower j3 to 0, and we shall need onlyj^ If m>0 and p<0, [6] will raise p to —1, and [4] will then /dx — z If m<0 andp<0, [6] will raise ^ to —1, and [3] will raise iTdx m to 0, and we shall need 1 — J xz f m 'dx 1 — = log a;, X fdx^ r J z J a Cdx^ r J xz J a. = -\og(a + bx), + bx b dx 1 , a + bx = — - loo; —^ x{a + bx) a x Hence, when w = 1, and m and j) are integers, our reduction for- mulas alwa3'^s lead to the desired result. J xr(a Examples. dx b*i a-\-bx , W b^ , b {a-]-bx) a^ X a}x 2a^xr Sa-a^ 4: ax* (2) Consider the case where n=2, rewriting the reduction formulas to suit the case, and giving an exhaustive investi- gation. ^ ^ J (a + bx-y^~ + 4:b{a-\-b3^y 8ab{a-\-bx^) tan'-'aj-v -• 8{ab)i \a 56 INTEGKAL CAI.CULTJS. [Art. 66. CHAPTER VI. lEEATIONAL FORMS. 66. We have seen that algebraic polj'nomials and rational fractions can alwa3'S be integrated. When we come to irrational expressions, however, ver^- few forms are integrable, and most of these have to be rationalized by ingenious substitutions. If an algebraic function is irrational because of the presence of an expression of the first degree under the radical sign, it can be easily made rational. Let /(x, "v/ct + bx) be the function in question. Let z = -\/a + bx ; then 2" = « -|- bx. nz^~^dz = bdx, -, nz'^-^dz dx = ; 6 2" — a x = b Hence (f{x, ^a + bx) dx = y ( /[ ~^ , 2 j«"'^ dz, which is rational and can be treated b}- the methods of Chapter IV. Examples. (1) C^l^L±ldx = x + 4: ^x + 4 log {^x-1). J -^x —1 (2) C^iax + bydx = ^^ V(«^' + bY+^ ^ J a (m + n) (3) J{xy{x + a) 4- V(^' + «)]f?^ 2 u + 1 n + 1 Chap. VI.] IRRATIONAL FORMS. 57 67. A case not unlike the last is j f{x, ^c + Va + bx)dx. Let z = a/c + Va + 6a; ; Hence z'' = c + Va + bx, (s"— 0)™ = a -\-bx, (2"— c)™— « dx = i-ii ^ |/(x% "Vc + \/a + bx)dx mn rS (z" — c)"" — a ~\. „ x„_i ,,_i -'^z''-'^dz. (1) Find r (2) Find f Examples. xdx Vc + V« + 6a; dx A/l+Vl-a; 68. If the expression under the radical is of a higher degree than the first the function cannot in general be rationalized. The onl}' important exceptional case is where the function to be integi'ated is irrational bj' reason of containing the square root of a quantity of the second degree. Required | /(a;, Va -\-bx-\- ca^)dx. First Method. Let c be positive ; take out Vc as a factor, and the radical may be written V^l + Bx + a^. Let -y/A + Bx + a;- = a; -|- 2, A + Bx -\- a^ =: x^ + 2xz + z^, z^-A . X = , B-2z 58 INTEGRAL CALCULUS. [Art. 68. a^^ 2{z'-Bz + A)dz {B-2zy ' J A . T, — ; — , z^ — Bz -\-A yA -\-Bx + x' = x-\-z = — -2— , B~2z and the substitution of these vahies will render the given func- tion rational. Second Method. Let c be positive ; take out Vc as a factor, and, as before, the radical may be written ^1 A + Bx + a?. Let sfA+Bx^^^ = -sj A + xz ; A +Bx -j-af = A-{-2 ^A . xz + a?z^, 2^A.z-B ''- \-z' ' . 2{^A.z'-Bz + yJA)dz ^ A+Bx + x^ =-^A + xz= ^^-''-^'^'^^ , 1 — Z" and the substitution of these values will render the given func- tion rational. If c is negative the radical can be reduced to the form ylA+Bx — a^, and the method just given will present no difficulty. Third Method. Let c be positive ; the radical will reduce to 'sJA -\- Bx + XT. Resolve the quantity under the radical into the product of two binomial factors {x — a)(x — (3), a and fi being the roots of the equation A + Bx -\-x^ = 0. Let V(a; — a) (x — yS) = (x — a)z ; (X — a) (x — /3) = (x — aYz^, X = ' z- a^^ 2z(f3-a)dz {1-zr ' ^/(x-a)(x-l3) = {x-a)z^ -^^^^^, 1 ^ Chap. VI.] lERATlONAL FOEMS. 59 and the substitution of tliese values will make the given function rational. If c is negative the radical will reduce to V^l + Bx — a^, and may be written ^f {a — x) {x — P) where a and ^ are the roots of x^ — Bx — xl = 0, and the method just explained will apply.. In general, that one of the three methods is preferable which will avoid introducing imaginary constants ; the first, if c > ; a a the second, if c < and > ; the third, if c < and — < 0. ' — c ' — c a If the roots a and (3 are imaginary, and A = ^^ is negative, it will be impossible to avoid imaginaries, for in that case A + Bx — x^ will be negative for all real values of x. 69. Let us compare the working of the three methods just /clx ' 1st. Let '\/'2-{-ox + x^ = x-{-z; r dx ^ r 2(z--Sz + 2)dz 3-22 ^ r 2dz J -v/9-u::^^.4-^^' J (3-2zY 'z'-3z + 2 J S-2z f ^2-{-dx + x^ ^ {S-2zr 2^-32 + 2 = -log(3-20), . = — log(3 + 2 X — 2 V2 + 3 ic + af') dx V2 + 3 a; + ar T 1 = log 3 + 2 a.' — 2 V2 + 3 a; + a;^ _, 3+2a; + 2V2+3a; + a^' ~ ^°9 + 12a; + 4.^•^-8 -12a;-4a;2 = log[3+2a; + 2V2 + 3a; + a;2]. (1) 2d. Let V2 + 3 a; + a;- = V2 + a;z ; r dx ^ 2 r(V2.^--3g + V2)c?2 ^ 1-z^ J V2 + 3a; + ar^ J (1-^')' ^2 .z'-Sz-{-'^2 = 2 C-^ = log^. (Art. 52) J 1 — Z' 1 — z 60 ■ INTEGRAL CALCULUS. [Art. 69. dx ^ j^^. x-^2 + V¥-\-3x + :x^ 's/2 + 'dx + x^ ° x + ^2 — ^2+3x-{-x^ ^» x'+2-^2.x + 2-2-nx-a^ \ S + 2x + 2^2 + 'dx-^x^ = ^"S 2Vi^3 = log(3 + 2x+2^2+-dx+x^) - log(2V2-3), or, di'opping the constant log (2 a/2 — 3) , r ^^^ — = log(3 + 2a; + 2V2 + 3a; + a^)- (2) •^ V2 + 3 a; + a;- 3d. Let V2 4-3iK + ar^= V(a^+1) {x + 2) = (x + l)z ', J 's/2 + Sx-\-x' J {1-zy -z Jl-z^ 1- z J -.n. dx 1+ f^+2 ^•^ + 1 _ i^^Va;+l + Va; + 2 : = log ,' = log V2 + 3a; + a.'^ °l_ !•£+_£ '^ Va;+ 1 — Va; +2 \a; + l , CK + 1 + 2 V2 + 3 a; + ic^ _^ a; _|_ 2 = loa; — ■ ■ ■ ■ ■ — ^ a; + l-a;-2 = log (3 + 2a^ + 2 V2 + 3a;+a^) + log (- 1), or, dropping the imaginarj" constant log ( — 1 ) , f ^^^ — =log(3 + 2a; + 2V2 + 3a; + a.-^;). (3) -^ V2 + 3x + «2 ' Examples. /dx n V4 + 2 ic — V2 — X ^^=r = 4 log — =1=^ • (2 + 3a;)V4-x2 V4 + 2a;+V2-a; (2^ r /^ = log (I- + X + Va^ + a.0 . J V a;- + a; (3) r , '^"^ - = — log f— + x^/c +Va + to + ca.-^\ ^ ■ J Va + to + cx- Vc \2Vc / Chap. VI.] IRRATIONAL FORMS. 61 70. If the function is irrational tlirough tlie presence, under the radical sign, of a fraction whose numerator and denominator are of the first degree, it can alwa^'s be rationalized. Required i/f .r, -T " "*" ) clx. J \ \lx-\-m Let z = ^'^^^±^ yilx+ m + m ^„ _ ax-\-b Ix -\- m X — , Iz'' — a I ,_ njcim — bl)z''~^dz and the substitution of these values will make the given function rational. Example. r dx sl l-x ^ _ 3 3 1/ 1 -a^ Y 71. If the function to be integrated is of the formx"'~\a-j-bx^y, m, ri, and j9 being an}' numbers positive or negative, and one at least of them being fractional, the reduction formulas of Art. 64 will often lead to the desired integral. Examples. = fsin ^x. (3 + 2ar), {l-x^)h dx 11 1 - Vl - x^ Vl ar'Vi_a;2 x 2ar' J {2ax — x-')i \2 2/ \2a C4^ f xhlx ^ {2a^ + Sx^) 62 INTEGRAL CALCULUS. [Art. 72, 72. We have said that when an irrational function contains a quantit}" of a higher degi-ee than the second, under the radical sign, it cannot ordinarily be integrated. It would be more cor- rect to saj- that its integral cannot ordinarily be finitely expressed in terms of the functions with which we are familiar. The integrals of a large class of such irrational expressions have been speciall}- studied under the name of Elliptic Functions. The}- have peculiar properties, and can be expressed in terms of ordinar}^ functions only by the aid of infinite series. Chap. VII.] TRANSCENDENTAL FUNCTIONS. 6'^ CHAPTER VII. TRANSCENDENTAL FUNCTIONS. 73 . In dealing with the integration of transcendental functions the method of integration by parts is general!}" the most effective. For example. Required | a; (log cc)- dec. Let M = (loga7)^, dv = x.dx ; '2,\ogx.dx du = » X a? p (log x) 2 = ^""(^^^^y _ Cx log x.dx = - [ (log x) 2 - log a; + i] . Again. Required i e'sm.x.dx. u = sin X, dv = e" dx ; dti = cos x.dx, I e^'sin x.dx = e^^sin a? — I e'cosx.dx, I e" cos x.dx = e' cos x-\- i e" sin x.dx ; , Ct ' 1 e''(sina; — coso;) whence I e* sm x.dx = -^ ' , , Ct 1 e*(sincc + cosx) and I e"" cos a;.ax'= — ^^ ■ '— J 2 64 INTEGRAL, CALCULUS. [Akt. 74. Examples. (1) x^'^ilogxydx = — — (log.'c)^- ^ »/ 6 loo; a; 6 (?ft + 1)- (m + 1)3 ^ ^ J (l-a;)2 1-a; ^"^ ^ (3) fe"^ V(l — e-'^).dx = J-e«== V(l — 6)^'"+ isin-^e<^. 74. The method of integration b}^ parts gives us important reduction formulas for transcendental functions. Let us con- sider I sin^x.dx. u = sin"~^a;, dv = sinx.dx; du = {n — l)sin'^~^xcosx.dx, V = ~ cos x ; I siu"a.'.cZa; = — sin"~^ ic cos cc + (n — 1) | sin"~-a; eos'X.dx = — sin"~^cccos"iK + (?i — 1) I (sin"~^a;— sin"a;)da; ; /s'lrf-x.dx = sin"~^x' cosa^ + ^^~ ) sin"+^a;.da;. [1] n n J Transposing, and changing n into n + 2, we get /I 7^ _f_9 /• sin" x.dx= sin" + ^ cc cos a; H ^^^ i sin" '^'^x.dx. r2 1 1i-t-l 71+1 J "- -^ In like manner we get / cos" x\c7ic= -since cos""' a^H ^^— | cos"~^a;.da;, [3] n n J /cos'^ x.dx = — — — ■sina.'cos"+'a; + ^^— t^ j cos" + ^a;.cZa;. r4'l 71 + 1 n + lJ ^ -^ If n is a positive integer, foi^mulas [1] and [3] will enable us to reduce the exponent of the sine or cosine to one or to zero, Chap. VII.] TEANSCEKDENTAL FUNCTIONS. 65 and then we can integrate by inspection. If n is a negative integer, formulas [2] and [4] will enable us to raise the ex- ponent to zero or to minus one. In the latter case we shall need , or I — — , which have been found in Art. 54 (c) . coscc J since Examples. 3 ■X. (1) Jsin^a;.dx = - ^^^Y^^^Ain^g; +1) + : /o\ C & J since cos^a^/ « , 5\ , 5 / . , ^ (2) I cos''a;.aa; = ( cos"a; + - H (smcccosce+a;). ^ ^ J 6 V 4y 16 ^ ^ (3) r-^ = -H^?^+ilogtanf. (4) Obtain the formulas fsh" x.dx = - Sh»-ia; Chcc - ^^—1 C^lx^-'^x.dx. [11 J n 71 J -^ rSh"a;.da; = ^— Sh"+ia;Chcc-^i±^ C^'\x''+^x.dx. [21 J n+l n + lj ■- -■ fch" x.dx = -^\xx Ch"-i X + ^^^ fch^-'ce.da;. [3] fch" x.dx ■-= ^— Sh X Ch" + ^ a; + ^^ fch" + ^ x.dx. [41 c/ 9i+l w+lJ ^ -■ ^ ^ J Sh'^o." 'Sh-x' * ^^ChcK+l 75. The (sin"^cc)"daj can be integrated by the aid of a reduc- tion formula. Let z — sin~^cc ; then cc = sin2:, dx^= cosz.dz, and I (sin~^a;)''da;= j 2"cos2.d2. 6Q INTEGRAL CALCULUS. [Art. 76. Let ?« = 2", dv = cosz.dz ; du = nz^'^dz, v = smz; i z"cosz.dz = z'"smz — ni z"~^sinz.dz. i z^'^sinz.dz can be reduced in the same waj^, and is equal to —z^~^cosz-\-{n — l) i z'^~^ cosz.dz; hence i z"" cosz.dz = z"smz + 'nz'"~'^ cosz — n(n — l) ( z"~^cosz.dz, [1] or I (sin~-^a;)"da; = cc(sin~^a;)"+ izVl — a3^(sin~^cu)"~^ — n(n — l) r(sin-^a;)"-^da;. [2] If w is a positive integer, this will enable us to make our re- quired integral depend upon | dx or | sin~^x.dx, the latter of which forms has been found in (I. Art. 81). Examples. (1) Obtain a formula for | (vers"^a;)"da;. (2) r(sin-^a;)*da; = a;[(sin-iaj)*- 4.3. (sin-^a^)2+4 .3.2.1] + 4 Vl - a^sin-^o; [(sin-^aj)2- 3 . 2]. 76. Integratioh by substitution is sometimes a valuable method in dealing with transcendental forms, and in the case of the trigo- nometric functions often enables us to reduce the given form to an algebraic one. Let it be required to find | (/;sina;) cosx.dx. Let 2; = sinx, dz = cosx.dx; I (fsmx) cosx.dx = ifz.dz. Chap. VII.] TRANSCENDENTAL FUNCTIONS. 67 In the same way we see that I {fcosx)smx.dx =— ifz.dz if z = cobx^ and j [/(since, cos cT)] cos a. da; = i [/(g, \J\—z^)'\dz if 2 = sin a;, I [/(sinx, cosa;)] sina;.c?x = — | \_f{z, '\Jl—z^)']dz if 2: = cosa:. 77. I sin'^a; cos^x.da; can be readily found by the method of Art. 76 if m and n are positive integers, and if either of them is odd. Let n be odd, then 71-1 cos" a; = cos""\r cos a; = (1 — sin^a;) 2 cos a;, I sin™ a; cos"a;.cLT = | sin™ a; (1 — sin'a;)^" cosaj.da;. Let 2 = sin a;, dz = co&x.dx, sin™ a.' cos" a;. da; = I 2™(1 — 2;^) 2 ^z^ which can be expanded into an algebraic polynomial and inte- grated directty. If m and n are positive integers, and are both even, I sin™ a; cos"a;.cZa; = | sin™a;(l — sin^a;)2'da;. n sin™a;(l — sin^a;)2" can be expanded and thus integrated by Art. 74 [1]. If m or n is negative, and odd, we can write cos" X = cos"~^ X cos X, or sin™ x = sin™"^ x sin a;, and reduce the function to be integrated to a rational fraction b}' the substitution of 2; = cosa;, or 2: = sina;. I sin™ a; cos" a;. da; can also be treated by the aid of reduction formulas easily obtained. 68 INTEGEAL CALCULUS. [Art. 77, I cos^ X Vsin x.dx-= Examples. , _ cos^^cc cos® a 10 8 2 sini cc _ 2 sin J x ~~i 7 rsin^^2cosi^_2c^g.^, »^ Vcosa; 5 /, -47 sin a; cos cc /sin* .T sin^cc 1\ , x cos-o. sm^x.do; = ^— (^-^^ 12- - 8 j + 16 — = seca? +log tan — sin X cos- X 2 /dx cos a:; , 3 , , x = sec X — \- - log tan- sin^ a; cos^ a? 2 sin" a; 2 2 /' dx 1,1,,. = — I -— — f- log sin a;, tan^a; 4 tan* a; 2 tan'' a; Chap. VIII.] DEFINITE INTEGRALS. 69 CHAPTER VIII. DEFINITE INTEGRALS. 78. A definite integral has been defined as the limit of a sum of infinitesimals, and we have proved that if the function to be integrated is continuous between the values between which the sum is to be taken, this limit can be found by taking the differ- ence between two values of an indefinite integral. In some cases it is possible to find the value of a definite integral from elementary considerations without using the indef- inite integral, and it is worth while to take one or two examples where this can be done. JIT cos^x.dx. By our definition this must equal limit [cos^O.daj + co&^ dx.dx -\- cos^2 dx.dx -{■ cos^Sdx.dx -\- -f- cos^(Tr—ddx) . dx + cos^(7r— 2 dx) . dx + cos^(7r— da;) . dx'] = limit [^dx + cos^ dx.dx + cos^2dx.dx + cos^Sdx.dx -j- — cos^3 dx.dx — cos^ 2 dx.dx — cos^ dx.dx^ , since cos (tt — ^) = — cos <^. We see that in this sum the terms destroy each other in pairs, with the exception of the first term dx if the number of terms is odd, and with the exception of the first term and a term cos^- dx.dx in the middle of the set, if n is even. Each of these 2 terms has zero for its limit as dx approaches zero ; hence cos^x.dx = 0. /' 70 mTEGKAL CALCULUS. [Art. 79. sirr'x.dx. XTT sin^x.dx = limfsiu^O.cZrK + sin^dx.dx + siTo?2dx.dx + sin^Sdx.dx + + sin^ (tt — 3 dx) . dx + sin^ (tt — 2 da;) . dx + sin^ (tt — dx) . da;] = lim[0 + 2s[n^dx.dx + 2 sin^ 2 dx.dx-\- 2 sin^ 3 dx.dx -\- ] = 2 1 ^sin-a;.da; = 21ini[sin^0.da; + siii^cZa;.cZa;+sin^2cZa;.da;+ + sin^ ( - — 2 dx \.dx + sin^ (- — dx\. dx"] = 21ini[sin^0.cto + sin^cZa;.da; + sin^2da;.da; + sin^3da;.da; + + cos^ 3 dx.dx + cos^ 2 dx.dx + cos^ da;, da;] = 2 lim [da; + da; + da; + ] . Since sin^ <^ + cos^ dx = a cos (fi.d(ji, dy — — b sin (f>.dcf), d^ = (a^cos^ <^ + b'sm^)d(lP = [ci^- (a^ - b') sin^ ']dcf>^ = cr-fl - ^!!^sinVy?<^' = a\l- e'sm'cf,)dcf>\ where e is the eccentricity of the ellipse. }' (I. Art. 150), Chap. IX.] LENGTHS OF CUKVES. 87 s — ai {l — e^sm^cf>)hd XI 2 ^24 ^ 2 4 6 r J '^ This is one of the famous Elliptic Integrals. Example. Show that the length of the hyperbolic arc is .=.£[i+fs.'. 'd = a — J r rd= adr, d^ = dr^-{- r^dcf>^ = (1 + a^)dr^ ; s = f(l + a')idr = (1+ a')h(r, - n). Examples. (1) Find the length of an arc of the parabola from its polar equation 1 + cos <^ (2) Find the length of an arc of the Spiral of Archimedes r = a(ji. 100. To rectify the Oardioide. "We have r=2a(l-cos<^), (I. Art. 109, Ex. 1), dr = 2asmcf).dcfi, d^z= 4a-sin^<^.d^^ 4-4a^(l — cos^)^d<^^ = 8a^d^^(l — cos)hdc^=8a cos^^-cos^' •J'Po [_ 2 2_ = 1 6 a for the whole perimeter. Chap. IX.] LENGTHS OF CUKVES. 89 Involutes^' 101. If we can express the length of the arc of a given curve, measured from a fixed point, in terms of the coordinates of its variable extremity-, we can find the equation of the involute of the curve. We have found the equations of the evolute of y=fx in the form x' = X — pCOSv y' = y — psmv We have proved that tanv = tanr', dp nT' = ^ ds dx' ]' and that 1, COST = ds'} (I. Art. 91). (I. Ai-t. 95), (I. Art. 96) ; (Art. 89). Since tanv = tanT', v = t' or v=180° + t'. As normal and radius of curvature have opposite directions, we shall consider v = 180° + r'. Then Hence smi/ = — smr X =:X-{- p- y' = y+pTj' and cos v = — cos t' dx' dy^ 'ds'' Since dp = ds', P = s' + l where I is an arbitrary constant, x and y being the coordinates of any point of the involute, it is only necessary to eliminate x', y', and p hy combining equations (1) , (2), and (3) with the equa- tion of the evolute. As we are supposed to start with the equation of the evolute and work towards the equation of the involute, it will be more natural to accent the letters belonging to the latter curve instead (1) (2) (3) 90 INTEGRAL CALCTJLITS. [Art. 101. of those going with the former ; and our equations may be written ^ _ ^.' _i_ „' ^' . els ' X =x'-\- p'"-^ (4) y = y'-\-p <(iy cls (5) p'=s+L (6) To test our method, let us find the involute of the curve f = 21m {x — mY, (7) letting p'=s-\-m. We must first find s. Q 2 ydy = {x — mydx, 9m _ 4 (a; — my 9 m y dy = ^ '-'" ""' dx, 2x-\-m ds' = 3 m dx^, V3 /Q' = 5 + m = 1 /^^ 1 — I (2a; + m)5dx' = — (2a; + m)i— m, ■ (2a:; + m)i, , , 2x -\-m x = x -\ ■ — 3 y = y'+ 3^3m -m 4 (2a; + ffl-) (a? — m)^ 27m 2/ , x — m X = «-, 4 , ,2 4a;'2 2/ = - — (a; - m)2 = , 92/ 2/ £c = 3a;'+m, 4a;'2 2/ = -- 2/' Substituting in (7) the values of x and y just obtained, we have y'^ = 2mx' as the equations of the required involute. Chap. IX. J LENGTHS OF CXJIIVES. 91 Example. Find the involute of ay^ = x^. .102. The involute of the cycloid is easily found. Take equa- tions I. Art. 100 ((7), x= aO -\- a sin 6 y = —a + a cos 6 p' = S, a dx=-a{l-\-cos6)d9 =2acos^-clO, a a dy = — asin0.d$ = — 2asin-cos-c?^, 2 2 ds^ = 2 a^ dO'' ( 1 + cos ^) = 4 a2 dO'" cos^ ^, Let =^"X cos-dd =4asm-? iC = aj' + 4asin-cos- = ;«' + 2a sin^, 2 2 ' ?/ = ?/' + 4a sin^- = ?/' — 2a(l — cos^), x' = aO — a sin ^ y' =z a — a cos ^ a cycloid with its cusp at the summit of the given cycloid. Example. From the equations of a circle X = a cos (f> y = a sin (^ obtain the equations of the involute of the circle. Ans. x'= a (cos ^ + ^ sin <^) ) y'= a(sin (fi — (fi cos cji) 3 92 INTEGEAL CALCULUS. [Art. 103. Intrinsic Equation of a Curve. 103. An equation connecting the length of the arc, measured from a fixed point of any curve to a variable point, with the angle between the tangent at the fixed point and the tangent at the variable point, is the intrinsic equation of the curve. If the fixed point is the origin and the fixed tangent the axis of X, the variables in the intrinsic equation are s and r. We have already- such an equation for the catenary s = atanr, Art. 90 (3), [1] the origin being the lowest point of the curve. The intrinsic equation of a circle is obviously s = ar, [2] whatever origin we may take. The intrinsic equation of the tractrix is easily obtained. We have y = — a sinr, Art. 91 (1) , and s = alog-; Art. 91, Ex. 1, hence s = a log ( — esc t) where t is measured from the axis of X, and s is measured from the point where the curve crosses the axis of Y. As the curve is tangent to the axis of Y, we must replace t by t — 90°, and we ^^ s = logsecT ^c(l-cosA^^ byArt.95[l]; therefore s = iM^L±A) (i _ cos '' r] [1] a \ a-\-2b J "- "^ is the intrinsic equation of the epic^'cloid, with the citsjy as origin. If we take the origin at a vertex instead of at a cusp a 7r(a + 25) ,. T f-T , 2a hence s' = — ^ ' sin t' ; a a + 2b 4b(a-\-b) . a or s = — ^^ — ■ — ^sin T a a + 2 6 is the intrinsic equation of an epicycloid referred to a vertex. 94 INTEGEAL CALCULUS. [Art. 105. Example. Obtain the intrinsic equation of the hypoej^cloid in the forms 45(a— 6)/-, a s — — ^^ ^ 1 — cos «-26 '• <^>- 4 6 (a— 6) . a .^. s = — 5^ ^sin T. (2) a a-2b ^ ^ 105. The intrinsic equation of the Logarithmic Spiral is found without difficulty. We have r=&e«, (Art. 98), and s= Vl+a2(ri-ro). (Art. 99). If we measure the arc from the point where the spiral crosses the initial line, ro = b, aud we have ^ s = 6Vl +a-(e"-l). In polar coordinates t = ^ + e, and in this case e = tan~^ a ; if we measure our angle from the tangent at the beginning of the arc we must subtract e from the value just given, and we have or, more briefly, s = k{c^ —1), k and c being constants. 106. If we wish to get the intrinsic equation of a curve directly from the equation in rectangular coordinates, the following method wUl serve : Let the axis of X be tangent to the curve at the point we take as origin. tanT = ^; (1) ax and as the equation of the curve enables us to express y in terms of x^ (1) will give us x in terms of t, say x = Ft', Chap. IX.] then hence LENGTHS OF CURVES. dx = F't.cIt, ♦ dx dx = F't-, but — = cost; ds ds ds ds = sec tF't. dr. 95 divide hy ds ; (2) Integrating both members we shall have the requked intrinsic equation. For example, let us take a^= 2 my, which is tangent to the axis of X at the origin. 2xdx = 2 mdy, dx tanT = X m dx = m sec^- r.dr, dx _ ds COSt = m sec^ T — ds = m sec^T.dr, ^ (1) r dr m[ sinr , , , /tt , t\] , ^ = m — -=- — ^+logtan -+- +(7, J cos"^T 2 |_cos^T \4 2J_ ■ = 0; .-. (7=0; f logtan^^ + I^ . s = when t = ; sinr COS^T (2) Examples. (1) Devise a method when the curve is tangent to the axis of F, and applj' it to 2/^ = 2 mx. Q (2) Obtain the intrinsic equation of ?/^ = {x — my. 21 m (3) Obtain the intrinsic equation of the involute of a circle. (Art. 102, Ex.) 96 INTEGRAL, CALCULUS. [Art. 107. 107. The evohde or the involute of a curve is easily found from its intrinsic equation. If the curvature of tlie given curve decreases as we pass along the curve, p increases, and s' z= p — po. (I. Art. 96) . If the curvature increases, p decreases, and s' = po — p. Hence always s' = ±(p — po) j [1] ds P = dr (I. Arts. 86 and 90). We see from the figure that t' = t. Hence s' = ± \dTjr=T' \drjT=o_ or, as we shall write it for brevit}", , ds s = ± — d [2] 108. The evolute of the tractrix s = a log sec t is d loo; sec - dr — atanr, the catenary. The evolute of the circle s = ar is d s = a- di = 0, a point. Chap. IX.] LENGTHS OF CURVES. 97 The evolute of the c^'cloid s = 4a(l — cost) is ,^4^cKl-cosr) = 4asinT, cIt an equal cycloid, with its vertex at the origin. Examples. (1) Prove that the evolute of the logarithmic spiral is an equal logarithmic spiral. (2) Find the evolute of a parabola. (3) Find the evolute of the catenary. 109. The evolute of an epicycloid is a similar epicycloid, with each vertex at a cusp of the given curve. Take the equation 4b(a + b) /^ a \ a ^ -,/^. r,-i 5^ — ■ — ^ 1 — cos -T . Art. 104ril. For the evolute. dl 1 — cos ■ 4:b(a + b) V a + 2b a cIt Ab(a-{-b) . a _^-, s — !^ — ! — ^sm — ■ T. ril a + 2b a + 26 •- -^ The form of [1] is that of an epicycloid referred to a vertex as origin ; let us find a' and &', the radii of the fixed and rolling circles. 4 6'(a' + 5') . a ^^ — H — - sui a' a'-\-2b hence, Ab'(a' + b') ^ 4b (a + b) s = - .' ' - sin , r, by Art. 104 [2] ; a +26' b (a + b) a' a + 2 6 ' a'+2b' a + 26 98 INTEGRAL CALCFLUS. [Art. 110. Solving these equations, we get a' = b' = a + 2b ah a + 26' a' _ a h'~V and the radii of the fixed and rolling circles have the same ratio in the evolute as in the original epicj'cloid ; therefore the two curves are similar. Example. Show that the evolute of a hypocjxloid is a similar hypo- cycloid. 110. We have seen that in involute and evolute r has the same value ; that is, t = t'. If s' and t' refer to the evolute, and s and t to the involute, we have found that , ds or di ^ s' = I, I being a constant, dr' ^ the length of the radius of curvature at the origin. {s' + l)dT' = ds, s= C'(s'-tl)dT' Jo . is the equation of the involute. The involute of the catenary s = a tanr is, when Z = 0, s = a I tanT.cZr = a log sec t, the tractrix. Chap. IX.] LEN^GTHS OF CUEVES. 99 The involute of tlie cycloid s = 4 a sinr when Z = is s = 4a I sinr.fZr = 4a(l — cost), an equal cycloid referred to its cusp as origin. The involute of a C3'cloid referred to its cusp s =4a(l —cost) when Z = is s = 4a I (1 — cos T)cZr = 4a(T + sinr), a curve we have not studied. The involute of a circle s = clt when Z = is = a I txIt = — Jo i 111. While any given curve has but one evolute, it has an infinite number of involutes, since the equation of the involute s'= C\s-^l)dT contains an arbitrar}' constant I ; and the nature of the involute will in general be different for different values of I. If we form the involute of a given curve, taking a particular value for Z, and form the involute of this involute, taking the same value of Z, and so on indefinitel}', the curves obtained will con- tinuall}' approach the logarithmic spiral. Let s=/r (1) be the given curve. S =jr' {I +fr)dT = It +£fr.dT is the first involute ; is the second involute ; S=Zr+^ + ^+ +^-^+ P'/t-^t" (2) 2 3! n\ Jo ^ is the wth involute. X 100 INTEGRAL CALCULUS. [Art. 112. B}^ Maclaurin's Theorem, fr=f0 + r/'o +^/"0 +l!/'"0 + 2 ! o ! But s = wlien r = ; hence /o = 0, and 2 ! 3 ! -^ 2 3! 4! »/o 3 ! 4 ! ! 0-^ (71+1)! (w+2)!^(n + 3)! ^^ as n increases indefinite!}'' all the terms of (3) approach zero (I. Art. 133), and the limiting form of (2) is Sz=It -\ \- 2! 3! = ^A+I + Zl + Il + _i\ Vl2!^3! J s = Z(e^- 1) by I. Art. 133 [2], which is a logarithmic spiral. 112. The equation of a curve in rectangular coordinates is readily obtained from the intrinsic equation. Given , s=/r, we know that sin r = — , as clx . and cosT = -— J as hence clx = costcZs = cost/'t.cZt, dy = sin rds = sin t/'txIt, COSrf'T.dr I y = i sin t/'txIt Chap. IX.] LENGTHS OF CURVES. 101 The elimination of t between these equations will give us the equation of the curve in terms of a; and y. Let us appl}- this method to the catenary. s = atanr, ds = a sec^T. dr. C' 7 1 jl+sinr a; = a I secT.ar = a log \ -. — , Jo \1— SUIT y = a j sec T tan T.dr = a(secT — 1), — 1 + sinr sinr = 1 — sinr e<^ —1 _€«■ — e a secT = |-(e«-fe «), 2/ = 2 («" + «") - «' the equation of the catenary- referred to its lowest point as origin. Curves in Space. 113. The length of the arc of a curve of double curvature is the limit of the sum of the chords of smaller arcs into which the given arc ma}' be broken up, as the number of these smaller arcs is indefinitely increased. Let (x, y, 2) , (x-j- dx, y -\- Ay, z + Az) be the coordinates of the extremities of anj- one of the small arcs in question; cLx, Ay, Ae are infinitesimal ; ^dx^-\-/\y--\- Az- is the length of the chord of the arc. In dealing with the limit of the sum of these chords, scny one may be replaced b}' a quantity dif- fering from it b}^ infinitesimals of higher order than the first. VcZx"^ + dy^-\- dz^ is such a value ; 1 VcZic^ + dy^ + dz^. I =10 102 INTEGRAL CALCULUS. [Art. 113. Let us rectify the helix. x = a cos 6 1 y = asine -. (I. Art. 214.) z = kO dx = — asmd.cW, cly = a cos 6. dO, dz = M6, ds- = (cr-{-Ji')de% s = {a' + k') h C \l6 = V^?TF {$1 -Bo). Examples. (1) Find the length of the curve (y = -—,z = -^X Ans. s=:x-{-z-\-L (2) y = 2'\/ax — x,z = x — %^— Ans. s = x + y — z. Chap. X.] AREAS. 103 CHAPTER X. AREAS. 114. "We have found and used a formula for the area bounded b}' a given curve, the axis of X, and a pair of ordinates. A - I ydx. Po-F We can readilj' get this formula as a definite integral area in the figiu'e is the sum of the slices into which it is divided b}' the ordinates ; if Ax, the base of each slice, is indefinitely decreased, the slice is infinitesimal. The area of any slice differs from yAx by less than /1?/A.T, which is of the second order if Ax is the principal infini- tesimal. We have then The «o * Aa; ^ limit ="=^==1 . by I. Art.lGl Hence Xn Examples. 1 xdy is the area bounded by a curve, the axis of Y, and perpendiculars let fall from the ends of the bounding arc upon the axis of Y. (2) If the axes are inclined at the angle w, show that these formulas become A. = siu CO I ydx = sin w I xdy. 104 INTEGRAL CALCULUS. £Art. ILo. 115. In polar coordinates we can regard the area between two radii vectores and the curve as the limit of the sum of sectors. Tlie area in question is the sum of the smaller sectorial areas, any one of whicli differs from |-r^A^ b}- less than the difference between the two circular sectors ^(r + A?-)^A<^ and ^?"A<^; that is, by less tlian ?-A?-A^ + -^^ — ^ — ^, which is of the second order if A^ is the principal infinitesimal. Hence limit \~,^.Z.^U..~\ ^9. 116. Let us find the area between the catenar}^ the axis of X, the axis of Y, and any ordinate. but Hence Jx Q nx X _x yclx= - I (e« + e a)dx. A=- (e«— e a) ^ (e« - e'a) = s, A = as, by Art. 90. and the area in question is the length of the arc multiplied b}- tlie distance of the lowest point of the curve from the origin. 117. Let us find the area between the tractrix and the axis ofZ. We have dx = -^^ Va^ - y"'. (Art. 91.) y A= i ydx = — j dy\ld- — y'^. Chap. X.] ABEAS. 105 The area in question is A = -Jdy Va-- y'=^, which is the area of the quadrant of a circle with a as radius. Example. Give, by the aid of infinitesimals, a geometric proof of the result just obtained for the traetrix. 118. In the last section we found the area between a curve and its asymptote, and obtained a finite result. Of course this means that, as our second bounding ordinate recedes from tlie origin, the area in question, instead of increasing indefinite^, approaches a finite limit, which is the area obtained. Whether the area between a curve and its asymptote is finite or infinite will depend upon the nature of the curve. Let us find the area between an hyperbola and its asj^mptote. The equation of the h3-perbola referred to its as^-mptotes as axes IS 2 xy = a' + W Let o) be the angle between the asymptotes ; then A = sin CO I ydx = ■ sm w I — = oo . Jo 4 Jo X Take the curve y~x = 4a^(2a — x), or 2/' = 4 a- ; X anj' value of x will giA^e two values of y equal with opposite signs ; therefore the axis of x is an axis of symmetry of the curve. When x = 2a^ y — ; as x decreases, y increases ; and when x = 0, y= CO . If a; is negative, or greater than 2 a, y is imagi- nary. The shape of the curve is something like that in the 106 INTEGRAL CALCULUS. [Art. 119. figure, the axis of F being a,u as3'mptote. The area between the curve and the asymptote is then either A=2 { yclx or A=2 i xdy ; by the first formula, A =*«fV 2a — X X b}'^ the second, A = Ua^ Examples. dy dx — 4a^7r . = 4a^7r. (1) Find the area between the curve y'^{x^ + cr) = a^x^ and its asj'mptote y = a. Ans. A = 2a^. (2) Find the area between y(2a — x) = x^ and its asj'mptote a; = 2 a. Ans. A = 3 tvoc. (3) Find the area bounded b}' the curve y'^= — - — ' and n 'Y» its as^'mptote x^=a Ans. A = 2a'il-{-- 119. If the coordinates of the points of a curve are ex- pressed in terms of'an auxiliary variable, no new difficulty is presented. Take the case of the circle x^ + y^ = a^, which ma}^ be written x = a cos = ttci^. Chap. X.] AREAS. 107 Examples. (1) The whole area of an ellipse ~ , . ^ >• is irdb. y =0 sill cji ) (2) The area of an arch of the cjxloicl is S-n-a^. (3) The area of an arch of the companion to the cycloid X — a$, y z=a{\ — cos6) is 27ra^ 120. If we wish to find the area between two curves, or the area bounded b}' a closed curve, the altitude of our elementary rectangle is the difference between the two values of y, which correspond to a single value of a). If the area between two curves is required, we must find the abscissas of their points of intersection, and the}' will be our limits of integration ; if the whole area bounded by a closed curve is required, we must find the values of x belonging to the points of contact of tangents parallel to the axis of Y. Let us find the whole area of the curve 49i7 9 4 O 7 cry + b-x* = cro'x-, or a*y'' = b'-x\a^-x^). The curve is s}Tnmetrical with reference to the axis of X, and passes through the origin. It consists of two loops whose areas must be found separately. Let us find where the tangents are parallel to the axis of Y'. y=—x V a~ — a^, a- dy b a^ — 2.^•^ , -^ = -5 — z=^^ = tanr. dx a- ^cr-x- T — - when tanr = 00, that is, when x = ± a. A = 2-^ Cx Va^ - x\dx + 2 ^ Cx Va^ - x'.dx = % ab. ccJ-a a- Jo 108 INTEGRAL CALCULUS, [Art. 121. Again ; find the whole area of {y ~xy = o? — y?. y = x ± Va^ — x^, A = C(y' - y")dx = C2 Va- - x'. To find the limits of integration, we must see where t = -• dii '\ld^ — a? ^ X 1 , -^ = — = 00 when cc = ± a . dx Va2 - x^ A==2 r Va^ -x' = tJ —a TTtt^ Examples. (1) Find the area of the loop of the curve y^ = — ^^ — -^— — ^ • a — x Ans. 2ci'(l-- (2) Find the area between the curves ?/" — 4aa; = and ^-^ay = 0. ^^,^^_ 16a_^_ 3 (3) Find the area of a loop of a^y^ = a:*(a^ — x^) . Ans. — -. (4) Find the whole area of the curve 2 y^ (a- + a;-) - 4 ay {a' -x') + (a' -x^ = 0. A " (a 5V2' V 2 121. We have seen that in polar coordinates Let us tr}' one or two examples. (a) To find the whole area of a circle. The polar equation is r=a. A = l \ c(? d4> = 7ra^. Chap. X.] AREAS. 109 (b) To find the area of the carclioide r = 2 a(l — cos ^) . ^4 = 1 ( 4a-(l - co&cl,ydct> = 2a- i (1- 2cos)dcf>, A=Qa^7r. (c) To find the area between an arch of the epicycloid and the circumference of the fixed circle. a + 6 x = (a + b) cos 6 — b cos y = [a + b)sm6 — b sin -^^ 6 We can get the area bounded by two radii vectores and the arch in question, and subtract the area of the corresponding sector of the fixed circle. Changing to polar coordinates, X = r cos (f), y = r sin ^. ^Ye want \ Crd^. y tan d> = —i o , 7 , xdy— ydx se& (bdcb = — "^—^ — ; ar but, since x = r cos <^, sec (^ = - ; X , r^dcb xdy— ydx hence ~ = — ^ — -J- — • , and 0'^ d(f) = xdy — ydx ; dx = (a-\-b)f- sin 6 + sin ^^-t-^ q) cW, dy = (a + b) fcos - cos ^^-±-^ e\ dO. xdy - ydx = (a + b) (a + 2b)fl- cos - oXw = 7^dcf>. Our limits of integration are obviously and — -■ a no INTEGRAL CALCULUS. .25T [Art. 121. Hence A = i(a + b){a-{-2b)J^ « fl-cos'^6\cie, A = -{a + b){a + 2b), a is the area of the sector of the epic3'cloicl. Subtract the area of the cu'cular sector irab, and we get j^^bUsi±2b)^ a as the area in question. (f?) To find the area of a loop of the curve r^ = a- cos 2 <^. For any value of cj) the values of r are equal with opposite signs. Hence the origin is a centre. When^ = 0, r = ±a; as 6 increases, r decreases in length tUl ^ = - , when ?• = ; as soon as <^ > -, r is imagiuarv. If 4 4 " decreases from 0, r decreases in length until 0= — ^ when r = ; and when ^<-, r is imaginar3^ To get the area of a loop, then, we must integrate from (^= — jto 4> = y A = i Cr\lcf> = \ a? P cos 2 c^.dtj = ^'• Examples. m (1) Find the area of a sector of the parabola r = — ^ ^ 1 + cos (^ (2) Find the area of a loop of the curve r-cos (^ = a- sin 3 4>. . ■ 3 a- a- , -, Ans. log2. 4 2 '' (3) Find the whole area of the curve r = a (cos 2 (^ + sin 2 <^) . Ans. TTfr. (4) Find the area of a loop of the curve rcos^ = a cos 2 4>. Ans. /'2-|yt-. (5) Find the area between r =a(sec<^+tan(^) and its asymp- tote rcos^ = 2 a. Ans. | + 2]«^. Chap. X.] AREAS. Ill 122. When the equation of a curve is given in rectangular coordinates, we can often simplify the problem of finding its area b}' transforming to polar coordinates. For example, let us find the area of Transform to polar coordinates. 7-2 = 4 (a^ cos- 4> + b- sin^ ^) , A=2i {a- cos^
  • ) clef, = 2 TV {a^ + b'). Examples. (1) Find the area of a loop of the curve (a;- + y-y = 4: crx^y^. Ans. — -• (2) Find the whole area of the curve 1- + ^=— 1- + ^ a* b* c^Kcc Ans. — — (or + b") . (3) Find the area of a Ipop of the curve 2/^ — 3 axy + a;^ = 0. Ans. 123. The area between a curve and its evolute can easily be found from the intrinsic equation of the curve. It is easily seen that the area bounded by the radii of curvature at two points infinitely near, by the curve and by the evolute, dif- fers from \f?dT by an infinitesimal of higher order. The area bounded b}' two given radii vectores, the curve and the evolute, is then A Hh 112 INTEGRAL CA1,CULUS. [Art. 124. P = ds Hence •'■^^Y*. ch For example, the area between a cycloid and its evolute is "i/d(4asinT) Let cIt rdr. To = and t^ ="''S ^COS^rdr = 2'7ra^. Examples. (1) Find the area between a circle and its evolute. (2) Find the area between the circle and its involute. Holditch's Theorem. 124. If a line of fixed length move with its ends on any curve which is always concave toward it, the area between the curve and the locus of a given point of the moving line is equal to the area of an ellipse, of which the segments into which the line is divided b}^ the given point are the semi- axes. Let the figure represent the given curve, the locus of P, and the envelope of the moving line. Let^P=a and PB = h, and let CB = p, C being the point of contact of the moving line with its envelope. Let AB = a-\-b = c. Chap. X.J AREAS. 113 The area between the first curve and the second is the area between the first curve and the envelope, minus the area between the second curve and the envelope. Let 6 be the angle which the moving line makes at any instant with some fixed direction. Let the figure represent two near positions of the moving line ; A^, the angle between these posi- tions, being the principal in- finitesimal. PB = p, P'B' = p + Ap. The area PBB'P'P differs from ^p^dO by an infinitesi- mal of higher order than the first. ip^de is the area of PBMP, and differs from PP'NB' by less than the rectangle on Plf and PQ, which is of higher order than the first, by I. Art. L53. But PP'JSTB diff"ers from PP'B'B by less than the rectangle on BN and NB', which is of higher order than the first, since NB', which is less than PP'+ Ap, is infini- tesimal and A^ is infinitesimal. The area between the first curve and the envelope is then ^ I p-d6 ; or, since we can take PP'A'A just as well for our elementary area, ^ i {c — pydd. ^2- ^2 7r Hence -J I p-d$=^ I (c — p)' dO whence j pdd = (1) The area between the second curve and the envelope is i£(p - hfde. 114 INTEGRAL CALCULUS. [Art. 125. The area between the first curve and the second is then ,27r ^2w = bipd6-b^7r dd by (1), = '7rbc — b^Tr = 7rb(a + b) — b^TT, A = Tvab, (2) which is the area of an eUipse of which a and b are semi- axes. Q. E. D. Examples. (1) If a line of fixed length move with its extremities on two lines at right angles with each other, the area of the locus of a given point of the line is that of an ellipse on the segments of the line as semi- axes. (2) The result of (1) holds even when the fixed lines are not perpendicular. Areas by Double Integration. 125. If we choose to regard x and y as independent variables, we can find the area bounded by two given curves, y = fx and y = Fx, by a double integration. Suppose the area in question divided into slices b}- lines drawn parallel to the axis of Y, and these slices subdi- vided into parallelograms by lines drawn parallel to the axis of X. The area of any one of the small parallelograms is AyAx. If we keep X constant, and take the sum of these rectangles from y=fx to y = Fx, we shall get a result differing from the area of the corresponding slice by less than Chap. X.] AEEAS. 115 2 Aa;Ay, which is infinitesimal of the second order if Ax and Ay are of the first order. Hence I Ax.dy = Ax idy is the area of the slice in question. If now we take the limit of the sum of all these slices, choosing our initial and final values of x^ so that we shall include the whole area, we shall get the area required. Hence A= ( 1 ( dindx. = f"( f'«)' tJXg \y/fX. J In writing a double integral, the parentheses are usually omit- ted for the sake of conciseness, and this formula is given as \ I dydx, the order in which the integrations are to be performed being the same as if the parentheses were actually written. If we begin b}- keeping y constant, and integj-ating with respect to cc, we shall get the area of a slice formed by lines parallel to the axis of X, and we shall have to take the limit of the sum of these slices varying y in such a way as to include the whole area desu'ed. In that case we should use the formula j dxdy. 126. For example, let us find the area bounded by the para- bolas 2/^ = 4 ax and a? — 4: ay. The parabolas intersect at the origin and at the point (4 a, 4 a). I ^Jydx, or A=X Xdxdy; ia dy = V 4 ax ; /y2 4a ia I I diidx =1 V4 ace ]dx=. — a^ Jo J^ ^ Ja \ 4.a) 3 4a The second formula gives the same result. 116 INTEGEAL CALCTJLTJS. [Art. 127. Examples. (1) Find the area of a rectangle by double integration ; of a parallelogram ; of a triangle. (2) Find the area between the parabola y^ = ax and the circle (3) Find the whole area of the curve {y — mx — cy = a^ — x^ Ans. TTtt^ 127. If we use polar coordinates we can still find our areas by double integration. Let r =f(f> and r = Fcfi be two curves. Divide the area between them into slices by drawing radii vectores ; then subdivide these slices by drawing arcs of circles, with the origin as centre. Let P, with coordinates r and ^, be any point within the space whose area is sought. The curvilinear rectangle at P has the base rA<^ and the altitude Ar ; its area differs from r/\(jiAr by an infinitesi- mal of higher order than rA<^A?\ The area of any slice as aba'b' is I rAcfiClr, (f} and Acfi being constant, that is A<^ | rdr. The whole area, the limit of the sum of such slices is ^= I I rdrd^. (1) Or we may first sum our rectangles, keeping r unchanged, and we get as the area of efe'f rAr I d^ and -4=1 I rdcf>dr. (2) 'i-H . 118 INTEGRAL CALCULUS. [Art. 128. CHAPTER XI, AREAS OF SUKFACES. Surfaces of Revolution. 128. If a plane curve y =fx revolves about the axis of X, the area of the surface generated is the limit of the sum of the areas generated by the chords of the infinitesimal arcs into which the whole arc may be broken up. Each of these chords will generate the surface of the frustum of a cone of revolu- tion if it revolves completely around the axis ; and the area of the sm^face of a frus- tum of a cone of revolution is, by element- ar}' Geometry, one-half the sum of the cir- cumferences of the bases multiplied by the slant height. The frustum generated by the chord in the figure will have an area differing by infinitesimals of higher order from ir{y -\-y + Ay) As or from 27ryds. The area generated by any given arc is then Ax s = 2 77 I yds [1] If the arc revolVes through an angle B instead of making a complete revolution, the surface generated is S^B » \yds. [2] Example. Show that if the arc revolves about the axis of y, S = 27r I xds. Chap. XI.] AREAS OF SUEFACES. 119 129. To find the area of a cj^linder of revolution. Take the axis of the cylinder as the axis of X. Let a be the altitude and b the radius of the base of the cylinder. The equation of the revolving line is 2/ = 6; ds = VcZcc^ + dy^ = dx ; S=27r i ydx = 2 TTCib, or the product of the altitude by the circumference of the base. Again, let us find the surface of a zone. The equation of the generating circle is af -\-y^ = a^ ; adx ds = y X(j x^ S = 2'7r i adx =2 aiT(xi — Xo). If Xq = — a and ccj = a, Hence the surface of a zone is the altitude of the zone multi- plied by the circumference of a great circle, and the surface of a sphere is equal to the areas of four great circles. Again, take the surface generated by the revolution of a cycloid about its base. x=:a6 — a sin i y = a — a cos $ V ds = adO V2 ( 1 - cos 6) , by Art. 94 ; aS = 2 TT raV2 . (1 - cos 0)i dd = ^iro?. 120 IKTEGEAL CALCULUS. [Art. 130. Examples. (1) The area of the surface generated by the revolution of the ellipse ^4-^=1 o? y" about the axis of X is 2 ircib J Vl — e^ + ^ j ; about the axis of Fis 27raM 1 + ~ ^ log —^\ V 2 e 1—eJ where e^= — . (2) Find the area of the surface generated by the revolution of the catenarj' about the axis of X ; about the axis of Y. (3) The whole surface generated by the revolution of the tractrix about its asj'mptote is 47ra^. (4) The area generated b}' the revolution of a cj'cloid about its vertical axis is 8 -n-a^^Tr — |) . (5) The area generated b}' the revolution of a cycloid about the tkngent at its vertex is ^^--n-a^. 130. If we know the area generated by the revolution of a curve about an}' axis, we can get the area generated b}- the revolution about any parallel axis by an easy transformation of coordinates. Given the surface generated by the arc from Sq to Si about OX, to find the area generated by the same arc when it revolves -x' about OX'. Let S be the surface about OX, and S' about O'X'. We have Po /S = 2 TT Cycls, S'= 2 TT Cy'ds'. Chap. XI.] AREAS OF SURFACES. 121 By Anal. Geom., x = x', y = yo + y'- Hence clx = dx', dy == dy', els = ds', and /S = 2 TT C(yo + y')ds = 2 7ri/o(si - Sq) + 2-nr Cy^'ds, = 27ryo{Si — So)-\-S'. Therefore ^' = ^S _ 2 7r?/o(si - Sq) . [1] Si — So is the length of the revolving curve ; 2 ttt/o is the cir- cumference of a circle of which y^ is the radius. Hence the new area is equal to the old area minus the area of a cj'linder whose length is the length of the given arc and whose base is a circle of which the distance between the two lines is radius. In using this principle careful attention must be paid to the sign of i/o, and it must be noted that the original formula S = = 2 TT j yds will alwa3's give a negative value for the area of the surface generated, if the revolving arc starts from below the axis ; and hence, that the surface generated by the revolution of any curve about an axis of sj'mmetry will come out zero. As an example of the use of the princi- ple, let us find the surface of a ring. Let a be the distance of the centre of — the ring from the axis, and b the radius of the circle. Since the area generated by the revolution of the circle about a diameter is zero, the required area is 2Trb.27ra = 4:Trhib. Example. Find the area of the ring generated by the revolution of a cycloid about any axis parallel to its base. Ans. S = AabTr[Tr-\ ■ ). 122 INTEGRAL CALCULUS. [Art. 131. 131. If we use polar coordinates, jS =27r I yds becomes S = 2 it \ r sm^.ds. where ds = Vc?r^ + tM^^. For example ; let us find the area of the surface generated by the revolution of the upper half of a cardioide about the hori- zontal axis. ?•= 2a(l— cos^) ; dr = 2asin<^.d<^, ds^ = 8 a^ ( 1 — cos <^) d<^^, ^ = 27r j 4 V2a2(l- cos<^)'siB«^.d(^. EXAIVIPLES. (1) Find the surface of a sphere from the polar equation. (2) Find the surface of a paraboloid of revolution from the polar equation of the parabola m r = 1 — cos ^ Any Surface. 132. Let X, y, z be the coordinates of an}^ point P of the sur- face, and X + Ax, y + Ay, z-\- Az the coordinates of a second point Q infinitely near the first. Draw planes through P and Q parallel to the planes of XY" and YZ. These planes will inter- cept a curved quadrilateral PQ on the sm-face ; its projection pg, a rectangle, on the plane of XZ\ and a parallelogram p'q' not shown in the figure on the tangent plane at P, of which pq is Chap. XI.] AREAS OF SUEFACES. 123 the projection. PQ will differ from p'q' by an infinitesimal of higher order, and therefore our required surface will be the limit of the sum of the pai'allelograms of which p'q' is any one. If (3 is the angle the tangent plane at P makes with XZ, 2y'q' cos ft = 2)q or p'q' =2)q sec ft. = Ax Az sec (3, and o-, our sur- face required, is equal to the double integral o- = | | sec ftdxd2 taken between limits so chosen as to embrace the whole surface. The equation of the tangent plane is {x-x,)D^J+{y-y,)DyJ+{z-z,)D,J=0, by I. Art. 217, (a;oi2/oi^o) standing for the coordinates of the point of contact, and f{x.,y,z) = being the equation of the surface. The direction cosines of the perpendicular from the origin upon the plane are D.J cos a = , „ : ? DyJ '^{D.jy + iDyjY + iD.jY DzJ COS v ^= — - ■ 5 ^ '^{D.Jf + iDyjy + iD.JY by Anal. Geom. of Three Dimensions. Q,osft= 124 INTEGRAL CALCULUS. [Art. 132. Hence, dropping the accents, By considering the projections upon the other coordinate planes we shall find ,^^ j^(Djy- + WY+iI>.f)\ yay_ 1-3] In each of the formulas the derivatives are partial derivatives. Let us find the area of the portion of the surface of the sphere a? -\- 'if -\- z^ = Cf? intercepted by the three coordinate planes. ^{Djy + {Djy+ (Djy = 2a. • o-= I I -dydz; (1) Jo Jox or o-= f \-dzdx; (2) Jo Joy or o-= \ \-dxdy. (3) Jo Joz For, in the second one, which agrees best with the figure, we must take our limits so that the limit of the sum of the projec- tions may be the quadi-ant in which the sphere is cut by the Chap. XI.] AEEAS OF SURFACES. 125 plane XZ ; and the equation of this section is obtained hj letting 2/ = in the equation of the sphere, and is whence z = Va^ — af. If we take as our limits in the integral ( - clz zero and -s/ar— x^ ^ y we shall get the area whose projection is a strip running from the axis of Z to the curve ; then, taking j ( j - f^ ) f?^ from to a, we shall get the area whose projection is the sum of all these strips, and that is om- required surface. y = Va^_a;2. (T=al I — Jo Jo Va^ -x'-z^ f clz Vci^ — 3Cp — = sin"-^ - Vcr — af if we regard x as constant ; c7o- >/rt2-a;2 dz cr — a \ -ax = — , Jo 2 2 the required area. Formulas (1) and (3) give the same result. 133. Suppose two cylinders of revolution drawn tangent to each other, and perpendicular to the plane of a great circle of a sphere, each having the radius of the great circle as a diameter ; required the surface of the sphere not included by the cylinders. The surface required is eight times the surface of which the shaded portion of the figure is the projection. If we take the plane of the great cu'cle as the plane of XY, 126 INTEGRAL CALCULUS. Q(? — ax + y- = (d is the equation of the cylinder, and o(? -\- 'if -\- z^ = a? [Art. 133. (1) of the sphere. We have o- = I 1 ■ From (2) (2) DJ=2x, {Djy+{Djy+{j)jy = ia\ dydx dydx. /» ra /• /» ayax Hence c.=J j J ^^^^- =' «J J v^^=^ r Our limits of integration for y are Vaa; — oi? and Va^ — a^; for X are and a. -^a-^-x-^clydx '^ax-x'^ J Vtt" — ce^ — — ain 1 2/^ Va^- = Sin" 9 '^l: a + ^ To find j sin-^-\U^ — dx we must integrate by parts. Jo \ a + £c Let and . _i I re w = sm ^A , \ a + a; dv = dx' ; du = --1 -.(Zic ; 2(a4-x) \x Jsin-^. a + cc ,dx = a sin \a-\-x "i J a+x Chap. XI.] AREAS OF SUEFACES. 127 Let to= -^x; 2 wcho — dx ,„a C:J^ = 2 f!^, = 2 ff 1 - -^)aw. '^"^^ J a + x J a + w" J\ a-\-io'J /'Jx.dx „/ , , _i w" -^ = 2[io — 'Ja tan ^ — = a + x \ ^ ^a Jsin'-^-y dx \a-\-x I- = asin"^-^ + «tan-U — a = — + — — a = — —a, 2 4 4 2 V2 2 ; 8cr = So? is the whole surface in question. Examples. (1) Find the area included by the cylinders described in Art. 133 b}' direct integration. (2) A square hole is cut through a sphere, the axis of the hole coinciding with a diameter of the sphere ; find the area of the surface removed. (3) A cj'linder is constructed on a single loop of the curve r=acos?i^, having its generating lines perpendicular to the plane of this cm've ; determine the area of the portion of the surface of the sphere a? +y--\-z' = a- which the cylinder inter- «ept«- Ans. ^(|-l). (4) The centre of a regular hexagon moves along a diameter of a given circle (radius = a) , the plane of the hexagon being perpendicular to this diameter, and its magnitude var3ing in such a manner that one of its diagonals alwaj's coincides with a chord of the cii'cle ; find the surface generated. Ans. a2(27r + 3V3). 128 INTEGRAL CALCULUS. [Art. 134. CHAPTER XII. VOLUMES. Single Integration. 134. If sections of a solid are made b}' parallel planes, and a set of cylinders drawn, each having for its base one of the sec- tions, and for its altitude the distance between two adjacent cutting planes, the limit of the sum of the volumes of these cj'linders, as the distance between the sections is indefinitely decreased, is the volume of the solid. We shall take as established b}' Geometrj^ the fact that the volume of a C3'linder or prism is the product of the area of its base b}' its altitude. It follows from what has just been said, that if, in a given solid, all of a set of parallel sections are equal, the volume of the solid is its base hy its altitude, no matter how irregular its form. Let us find the volume of a pjTamid having b J\ for the area of its base, and a for its altitude. // '\ Divide the pyramid b}' planes parallel to the // 'A base, and let z be the area of a section at the dis- /- -/ — i\ tance x from the vertex. , / y H We linow from Geometry that - = - . / / \ \ h cc \ I ■■•A TT & " V 'A Hence ^ = -<> ^- ' a- Let the distance between two adjacent sections be dx ; then the volume of the cylinder on z is and F, the required volume of the pyramid, is V=— \ x-dx = — . Chap. XII.] VOLUMES. 129 Precisel}' the same reasoning applies to any cone, which will therefore have for its volume one-third the product of its base by its altitude. Example. Find the volume of the frustum of a pyramid or of a cone. 135. If a line pKRce^ keeping always parallel to a given plane, and touching a plane curve and a straight line parallel to the plane of the curve, the surface generated is called a conoid. Let us find the volume of a conoid when the director line and curve are perpendicular to the given plane. Divide the conoid into laminae by planes parallel to the fixed plane. Let Ay be the distance between two adjacent sections, and let x be the length of the line in which any section cuts the base of the conoid ; let o be the altitude and h the area of the base of the figure. Any one of our elementary C3"linders will have for its volume |^aa;A?/, since the area of its triangular base is ^ax, and we have V—^a | xdy., the limits of integra- tion being so taken as to embrace the whole solid. | xdy be- tween the limits in question is the area of the base of the CO" noid ; hence its volume, EXABIPLES. (1) Find the volume of a conoid when the director line and curve are not perpendicular to the given plane. (2) A woodman fells a tree 2 feet in diameter, cutting half- way through from each side. The lower face of each cut is horizontal, and the upper face makes an angle of 45° with the horizontal. How much wood does he cut out? 130 INTEGEAI. CALCULUS. [Art. 136. 136. To find the volume of an ellipsoid. (J? h^ c^ Take the cutting planes parallel to the plane of XY". A sec- tion at the distance z from the origin will have c^ y^ _. _^'_ c^ — z^ a I h I for its equation, and — Vc^ — z"^ and - Vc^ — z^ for its semi-axes ; hence its area will be — — (c^ — z') . & Any of the elementary cylinders will have for its volume ^^(c^ — 2;^)A2:, and we shall have for the whole solid ■KCib If a, &, and c are equal, the ellipsoid is a sphere, and Examples. (1) Find the volume included between an hyperboloid of one sheet and its asymptotic cone ^ _l_^ ?- = 0. G? IP' (^ Ans. It is equal to a cylinder of the same altitude as the solid in question, and having for a base the section made by the plane of XY". (2) Find the whole volume of the solid bounded by the surface ^j^tj^t = \. STTCtbc r,2 ^ 7,2 ^ p4 Ans. — - — Chap. XII.] VOLUMES. 131 (3) Find the volume cut from the surface c b by a plane parallel to the plane of ( TZ) at a distance a from it. Ans. 7r«^^(6c), (4) Find the whole volume of the solid bounded by (ay^ -{- y^ + z-y = 27 a^xtjz. ■ A7is. fa^ (5) The centre of a regular hexagon moves along a diameter of a given circle (i-adius = a), the plane of the hexagon being perpendicular to this diameter, and its magnitude varying in such a manner that one of its diagonals alwaj's coincides with a chord of the circle ; find the volume generated. Ans. 2y3.a^. (6) A circle (radius = a) moves with its centre on the cir- cumference of an equal circle, and keeps parallel to a given plane which is perpendicular to the plane of the given circle ; find the volume of the solid it will generate, 9 3 A71S. f^(37r + 8). o Solids of Revolution. Single Integration. 137. If a solid is generated b}* the revolution of a plane curve y = fx about the axis of x,- sections made by planes perpendicu- lar to the axis are circles. The area of any such circle is Try-, the volume of the elementar}" cylinder is iry^Ax, and ' V = /^ X r I y^dx is the volume of the solid generated. For example ; let us find the volume of the solid generated by the revolution of one branch of the tractrix about the axis of X. Here we must integrate from a; = to a; = 00 . V y^dx. 132 INTEGRAL CALCULUS. [Art. 138. We have dx = - ^"^ ~ y''>" dy Art. 91 [2] y in the case of the traetrix ; hence V= ~ tc \ y{a^ — y^Y^dy. When a; = 0, y = a^ and when re = oo, y = 0. Therefore F= — tt I y{a^ — y^)idy = ttCV Examples. (1) Kthe plane curve revolves about the axis of Y, V—TT I x-dy. (2) The volume of a sphere is % -n-ce. (3) The volume of the solid formed b}" the revolution of a C3'cloid about its base is 5 tt^ cc^. (4) The curve y^{2a — x) = x^ revolves about its asymptote ; show that the volume generated is 2TT^a^. Solids of Revolution. Double Integration. 138. If we suppose the area of the revolving curve broken up into infinitesimal rectangles as in Art. 125, the element AxAy at any point P, whose coordinates are x and y, will generate a ring the volume of which will differ from 27ry^xAy by an amount which will be an infinitesimal of higher order tlian the second if we regard Ax and Ay as of the first order. For the ring in question is obviousl}^ greater than a prism having the same cross-section AxAy, and ha-ving an altitude equal to the inner circumference 2 Try of the ring, and is less than a prism having Ax Ay for its base and 27r{y -j- Ay), the outer circumfer- ence of the ring, for its altitude ; but these two prisms difier by 2irAx(Ayy, which is of the third order. Chap. XII.] VOLUMES. 133 Aa; I 2 irydy, where the upper limit of integration is the ordi- nate of the point of the curve immediately ahove P, and must be expressed in terms of x hy the aid of the equation of the revolv- ing curve, will give us the elementary cylinder used in Art. 137. The whole volume required will be the limit of the sum of these c^iinders ; that is, V=2-^r fydydx. [1] If the figure revolved is bounded by two curves, the required volume can be found by the formula just obtained, if the limits of integration are suitably' chosen. Let us consider the following example : A paraboloid of revolution has its axis coincident with the diameter of a sphere, and its ^-ertex in the surface of the sphere ; required the volume between the two surfaces. Let y^=2 7nx (1) be the parabola, and of -\-y^ — 2 ax = - (2) be the circle, which form the paraboloid and the sphere by then- devolution. The abscissas of their points of intersection are and 2 (a — 7n) . We have V=2Tri i ydydx, and, in performing our first integration, our limits must be the values of y obtained from equations (1) and (2). We get F= TT j [2 (a — m)x — xr']dx, and here our Imiits of integration are and 2 (a — m). Hence F= |7r(a — 7?i)^ = -— , 6 if 7i is the altitude of the solid in question. Examples. (1) A cone of revolution and a paraboloid of revolution have the same vertex and the same base ; required the volume be- tween them. ^^^_ ^rrf ^^^^^ j^ .^ ^^^ ^^^.^^^^ ^^ ^^^ ^^^^^_ 134 INTEGRAL CALCULUS, [Art. 139. (2) Find the volume included between a right cone, whose vertical angle is 60°, and a sphere of given radius touching it along a circle. j^^^^ ■^ 6 ■ Solids of Revolution. Polar Formula. 139. If we use polar coordinates, and suppose the revolving area broken up, as in Art. 127, into elements of which rclt^dr is the one at any point P whose coordinates are r and <^, the element rdcftdr will generate a ring whose volume will differ from 2777-^ sin (/)d(/)C?r by an infinitesimal of higher order than the second, if we regard dcfi and d?' as of the first order ; for it will be less than a prism having for its base rdcfidr, and for its alti- tude 2 IT (r + dr) sin {(f>-\-d(p), and greater than a prism having the same base and the altitude 2 Trr sin (ft ; and these prisms differ b}' an amount which is infinitesimal of higher order than the second. "We shall have then F= 2 TT rp-^ sin c^cZrdc^, [1] the limits being so taken as to bring in the whole of the gener- ating area. For example ; let us find the volume generated by the revolu- tion of a cardioide about its axis. ?'= 2a(l — cos^) is the equation of the cardioide ; ' F= 2 TT ffr^ sin <^dr#. Our first integral must be taken between the limits r = and r = 2 a (1 — cos 0) , and is (1— cos^)®sin<^d^. 3 16 r'^ '= — a^TT I (1— cos^)^sin(^(^d), 3 Jo 3 Chap. XII.] VOLUMES. 135 Example. A right cone has its vertex on the surface of a sphere, and its axis coincident with the diameter of the sphere passing through that point ; find the volxune common to the cone and the sphere. Volume of any Solid. Triple Integration. 140. If we suppose our sohd divided into parallelopipeds by planes parallel to the three coordinate planes, the elementary 31 parallelopiped at any point (x,y,z) within the solid will have for its volume AccA?/A2, or, if we regard x, y, and z as independent, dxdydz ; and the whole volume 17'= j I I dxdydz, [1] the limits being so chosen as to embrace the whole solid. The integrations are independent, and may be performed in any order if the limits are suitably chosen. For example ; let us find the volume of the portion of the ellipsoid ^2 ^2 ^2 cut off by the coordinate planes. 136 INTEGRAL CALCULUS. [Art. 140. 7"= j I I dzdydx, and our limits are, for z, and c\ll ^ — 4^; for y, and ^ \ a^ ¥ 6-*|l — — ; and for x, and a. For, starting at any point (x,y,z) and integrating on the h3'potliesis that z alone varies, we get a column of our elementarj' parallelopipeds having dxdy as a base and passing through the point {x,y,z). To make this col- umn reach from the plane XY to the surface, z must increase from the value zero to the value belonging to the point on the surface of the ellipsoid which has the coordinates x and y ; that is, to the value ca|1 5 — -^- Then, integrating on the h}'- \ cr b- pothesis that y alone varies, we shall sum these columns and shall get a slice of the solid passing through {x,y,z) and having the thickness, dx. To make this slice reach completely across the solid, we must let y increase from the value zero to the greatest value it can have in the slice in question ; that is, to the value which is the ordinate of that point of the section of the ellipsoid by the plane XF which has the abscissae. The section in question has the equation ~2 "■" 12 — ' therefore the required value of y is h >R- Last, in integrating on the hj^pothesis that x alone varies, we must choose our limits so as to include all the sUces just de- scribed, and must increase x from zero to a. /' &=.=caji-5-|; between the limits and cH 1 — — , — ^„ ^ a- W Chap. XII.] VOLUMES. 137 ^U-i-py 5/"s|»'('l-5)-/-'-!!/ nrhn. / . (rr\ ^ Ct — 1£^ fi _ ^ between the limits and 6-\ 1 . 4.1 V «V 6 ' the vokime required. Examples. (1) Find the vokime obtained in tlie present article, perform- ing the integrations in the order indicated b}' the formula, F'= I I I dxclydz. (2) Find the volume cut off from the surface t^y^^2x c^ b hj a plane parallel to that of TZ, at a distance a from it. Ans. •7ra^V(&c), (3) Find the volume enclosed by the surfaces, Ans. x^ -\-y^ = cz, 7? -\-y- = ate, 2 = 0. ^ -'■''' 32 c (4) Obtain the volume bounded by the surface 'Z = a — Va-^ + y^ and the planes x = z and x = 0. Ans. ^ 9 138 INTEGRAL CALCULUS. [Art. 141. 5. Find the volume of the conoid bounded by the surface 2 2 nrf^ ft 2 _^ ^ y _ g2 ^y^^ j^jjg planes x = and x = a. Ans. — -— . 141. If we use polar coordinates we can take as our element of volume an expression easily obtained from the element 2 -n-r^ sin cf>drd(f> used in Art. 139. Then V= C f Ci^ sin cfidrdcjide, where the order of the integrations is usually immaterial if the limits are property chosen. Example. Find the volume of a sphere by polar coordinates. Chap. XIII.] CENTRES OE GRAVITY. 139 CHAPTER XIII. CENTRES OF GRAVITY. 142. The moment of a force about an axis is the product of the magnitude of the force by the perpendicular distance of its line of direction from the axis, and measures the tendency of the force to produce rotation about the axis. The force exerted by gravity on any material body is propor- tional to the mass of the bod}', and may be measured by the mass of the body. The Centre of Gravity of a bod}' is a point so situated that the force of gravity produces no tendency in the bod}' to rotate about an}"" axis passing through this point. The subject of centres of gravit}- belongs to Mechanics, and we shall accept the definitions and principles just stated as data for mathematical work, without investigating the mechanical gTounds on which they rest. 143.. Suppose the points of a body referred to a set of three rectangular axes fixed in the body, and let x,y^z be the coordi- nates of the centre of gi-avit}-. Place the body with the axes of X and Z horizontal, and consider the tendenc}'' of the particles of the body to produce rotation about an axis through (x,y,z) parallel to OZ, under the influence of gravit}^ Represent the mass of an elementary paraUelopiped at an}- point {x,y,z) by dm. The force exerted by gravity on dm is measured b}' dm, and its line of direction is vertical. If the mass of dm were concen- trated at P, the moment of the force exerted on dm about the 140 INTEGRAL CALCULUS. [Art. 143. axis through C would be {x — x)dm, and this moment would represent the tendency of dm to rotate about the axis in ques- tion ; the tendenc}^ of the whole bod}' to rotate about this axis would be 1{x — x)dm. If now we decrease dm indefinitely, the error committed in assuming that the mass of dm is concentrated at P decreases indefinitel}', and we shall have as the true expres- sion for the tendenc}' of the whole bocl}^ to rotate about the axis through C, I (^ — x)dm ; but this must be zero. Hence i (x — x)dm = 0, I xdm — xi dm = 0, I xdm ^=7^ [1] I dm K we place the body so that the axes of Y and X are hori- zontal, the same reasoning will give us y = I ydm [2] I dm and in like manner we can get I zdm ^ = > [3] I dm Since i dm is the mass of the whole body, If we represent it by M we shall have ^ I xdm x = M J y = I ydm J M zdm M Chap. XIII.] CENTRES OF GRAVITY. 141 Example. Show that the effect of gravity in making a bod}' tend to rotate an}- given axis is precisely' the same as if the mass of the body were concentrated at its centre of gravity. 144. The mass of any homogeneous bod}' is the product of its volume b}' its density. If the bod}' is not homogeneous, the density at any point will be a function of the position of that point. Let us represent it by k. Then we may regard dm as equal to kcIv if civ is the element of volume, and we shall have I XKdv ^--c — [1] I Kdv and corresponding formulas for y and z. If the body considered is homogeneous, k is constant, and we shall have xdv i xdv I a [2] /.. y zdv [4] In any particular problem we have only to express dv in terms of the coordinates. Plane Area. 145. If we use rectangular coordinates, and are dealing with a plane area, where the weight is uniformly distributed, we have dv = dA = dxdy. (Art. 125) . 142 INTEGRAL CALCULUS. Hence, by 144, [2] and [3], j I xclxdy [Art. 145. and I I dxdy j j ydxdy , I I dxdy If we use polar coordinates, dv — dA = rd(f>dr, I I 9*^ cos (f> ddr [2] For example ; let us find the centre of gravity of the area be- tween the cissoid and its asj^mptote. From the equation of the cissoid r = » a — x we see that the curve is S3'mmetrical with respect to the axis of X, passes through the origin, and has the hne x = a as an asymptote. From the s}Tnmetry of the area in question, y.= 0, and we need only find x. ' I I xdydx I xydx - _ Jo J-y _ Jo I I dydx I ydx Jo J-y Jo Chap. XIII.] CENTRES OF GEAVITY. 143 Jo(a-xy^ _ Jo(«-x). . byArt.64[4]. Jo{a — x)^ Jo(a — a;)5 As an example of the use of the polar formulas [2], let us find the centre of gravity of the cardioide r= 2a(l— cos^). Here, from the fact that the axis of X is an axis of symmetry, we know that y = 0. i '1^ COS (J3drd I I rdrd^ i I r^co&4>d(f> —— I (1 — eosycoscf>d(f> _ Jo o Jo i r?-2 d 2a^ C{1- cos f dcji X27r (cos<^ — 3 cos^^ + 3 cos^c^— cos*^)dq!) = — -V"^ 5 X27r (1 — 2cos^ + cos-^)cZ0 = 37r. Hence x = — ^a. Examples. 1. Show that formulas [1] hold even when we use oblique coordinates. 2. Find the centre of gravit}^ of a segment of a parabola cut off by any chord. Ans. i=|a, y = 0. If the axes are the tangent parallel to the chord and the diameter bisecting the chord. 144 INTEGRAL CALCULUS. [Art. 146. 3. Find the centre of gravitj" of the area bounded b}' the semi- cubical parabola ay'^ = x^ and a double ordinate. Ans. x = -f-cc. 4. Find the centre of gravity of a semi-ellipse, the bisecting line being any diameter. Ans. If the bisecting diameter is taken as the axis of Y, and 4: (X the conjugate diameter as the axis of X, x = — , y = 0. 5. Find the centre of gravit}^ of the curve y"^ — b^ ~ . X Ans. x==^a. 6. Find the centre of gravity of the cycloid, Ans. X = air, y=^a. 7. Find the centre of gravity of the lemniscate ?- = a- cos 2 eft. . - 7rV2 Ans. x = a. 8 8. Find the centre of gravity of a circular sector. A71S. If we take the radius bisecting the sector as the axis of X, and represent the angle of the sector b}^ a, ^ = f a 9. Find the centime of gravity of the segment of an ellipse cut off b}' a quadrantal chord. Ans. x = ^ -, y = ^ -• TT 2 TT 2 10. Find the centre of gravity of a quadrant of the area of the curve oji -f 2/1 = al. Ans. x = y = fff -. TT 146. If we are dealing with a homogeneous solid formed by the revolution of a plane curve about the axis of X, we have civ = 2 irydydx. (Art. 1 38 [ 1 ] ) Hence, by Art. 144 [2], I j xydxdy I I ydxdy Chap. XIII.] CENTE-ES OF GEAVITY. 145 If we use polar coordinates, dv = 2 TTi^ sin cl>drdcf>. (Art. 139 [1] ) i i 'i^ sin (^ cos cf)drd(f> Hence x = '^ ^ [2] I 1 9-^ sin (^drdcji For example ; let us find the centre of gravit}^ of a hemisphere. The equation of the revolving curve is a? -{--if = a^. fxydydx ^ Jo If we use polar coordinates the equation of the revolving curve is r = a. 1 1 r^ sin rf) cos diddtdr , . Jo _i« . I I 1-^ sin x; the chance of this for any given value a-Q of x is — cos ^ — . The probability that x will have the value x^ is — . The probability required is iX . 2c * C X Jq c ■n-a This problem ma^j be solved by another method which pos- sesses considerable interest. Chap. XIV.] MEAN VALUE AND PEOBABILITY. 153 Since all values of x from to a, and all values of 6 from — | to - are equally probable, the whole number of cases that can arise ma}' be represented b}- P" Cdxde = TTCI. The number of favorable cases will be represented b}' \ .c COS 9 ( \ dxcW = 2i 2c Hence p = — TTtt (c) To find the probability that the distance of two stars, taken at random in the northern hemisphere, shall exceed 90°. Let a be the latitude of the first star. With the star as a pole, describe an arc of a great circle, dividing the hemisphere into two lunes ; the probability that the distance of the sec- ond star from the first will exceed 90° is the ratio of the lune not containing the first star to the hemisphere, and is equal to vr^LZL^. The probability that the latitude of the first star will be between a and a + da is the ratio of the area of the zone, whose bounding circles have the latitudes a and a -\- da. respectivel}', to the area of the hemisphere, and is 2 Tra- cos a da ■, = cos a aa. 2 TTcr a) , 1 Hence p= I "il^I — ^ cos a da •^0 TT (d) A random straight line meets a closed -convex curve ; what is the probability that it will meet a second closed convex curve within the first ? If an infinite number of random lines be drawn in a plane, all du-ections are equally probable ; and lines having any given 154: . INTEGEAL CALCULUS. [Art. 151. direction will be disposed with equal frequenc}' all over the plane. If we determine a line b}' its distance p from the origin, and b}' the angle a which p makes with the axis of X, we can get all the lines to be considered b}' making p and a vary between suitable limits b}' equal infinitesimal increments. In our problem, the whole number of lines meeting the exter- nal curve can be represented by | | clpda. If the origin is within the curve, the limits for p must be zero, and the perpen- dicular distance from the origin to a tangent to the curve ; and for a must be zero and 2 7r. If we call this number N, we shall have N = pda^ p being now the perpendicular from the origin to the tangent. If we regard the distance from a given point of any closed convex curve along the curve to the point of contact of a tan- gent, and then along the tangent to the foot of the perpendicu- lar let fall upon it from the origin, as a function of the a used above, its differential is easil}' seen to be pda. If we sum these differentials from a = to a = 2 tt, we shall get the perimeter of the given curve. Hence -ZV = i pda = L, where L is the perimeter of the curve in question. B}' the same reasoning, we can see that %, the number of the random lines which meet the ifiner curve, is equal to Z, its perimeter. For p, the required probability, we shall have I ^ = L Examples. (1) A number n is divided at random into two parts ; find the mean value of their product. ,^^^^ ^ 6* Chap. XIV.] MEAN VALUE AND PROBABILITY. 155 (2) Find the mean value of the ordinates of a semicircle, sup- posing the series of ordinates taken equidistant. Ans. -a. 4 (3) Find the mean A'alue of the ordinates of a semicircle, sup- posing the ordinates drawn through equidistant points on the circumference. a 2a TV (4) Find the mean values of the roots of the quadratic X- — ax -f- & = 0, the roots being known to be real, but b being unknown but positive. a , 5a -, a 6 " 6" (5) Prove that the mean of the radii vectores of an ellipse, the focus being the origin, is equal to half the minor axis wheh they are di'awn at equal angular intervals, and is equal to half the major axis when the}' are drawn so that the abscissas of their extremities increase uniformly. (6) Suppose a straight line divided at random into three parts ; find the mean value of their product. - a^ 60 " (7) Find the mean square of the distance of a point within a given square (side = 2 a) from the centre of the square. A71S. f a^. (8) A slab is sawed at random from a round log, find its mean tliickness. . 4 a Ans. — . o OTT (9) A chord is drawn joining two points taken at random on a circle ; find the mean area of the less of the two segments into which it divides the cii'cle. ^ 7ra^ a^' Ans. 4 TT (10) Find the mean latitude of all places north of the equator, Ans. 32°. 7. (11) Two points are taken at random in a triangle ; find the mean area of the triangular portion which the line joining them cuts off from the whole triangle. Ans. | of the whole. (12) Find the mean distance of points within a sphere from a given point of the surface. Ans. |a. 156 INTEGRAL CALCULUS. [Art. 151. (13) Find the mean distance of two points taken at random within a sphere. Ans. |fa. (14) Two points are taken at random in a given line a ; find the chance that their distance shall exceed a given value c. ^a — c\- Ans. a (15) Find the chance that the distance of two points within a square shall not exceed a side of the square. Ans. tt — -^^-. (16) A line crosses a circle at random ; find the chances that a point, taken at random within the circle, shall be distant from the line by less than the radius of the circle. . . 2 •^ Ans. 1 — 77-- OTT (17) A random straight line crosses a circle ; find the chance that two points, taken at random in the circle, shall lie on oppo- site sides of the line. . 128 Ans. 45 TT^ (18) A random straight line is drawn across a square ; find the chance that it intersects two opposite sides. loo- 2 ^ A71S. i ^. (19) Two arrows are sticking in a circular target; find the chance that then- distance apart is greater than the radius. Ans. 3V3 47r (20) From a point in the circumference of a circular field a projectile is thrown at random with a given velocity which is such that the diameter of the field is equal to the greatest range of the projectile ; ^nd the chance of its falling within the field. Ans. i--(V2-l). TT (21) On a table a series of equidistant parallel lines is drawn, and a cube is thrown at random on the table. Supposing that the diagonal of the cube is less than the distance between con- secutive straight lines, find the chance that the cube will rest without covering any part of the lines. Ans. 1 , where a is the edge of the cube and c the dis- ttC tance between consecutive lines. Chap. XIV.] MtlAN VALUE AND PROBABILITY. 157 (22) A plane area is ruled with equidistant parallel straight lines, the distance between consecutive lines being c. A closed curve, having no singular points, whose greatest diameter is less than c, is thi'own down on the area. Find the chance that the curve falls on one of the lines. Alls. — , where I is the perimeter of the curve. TTC (23) During a heavy rain-storm, a circular pond is formed in a circular field. K a man undertakes to cross the field in the dark, what is the chance that he will walk into the pond? 158 INTEGEAL CALCULIJS. [Art. 152. CHAPTER XV. KEY TO THE SOLUTION OF DIFFEEENTIAL EQUATIONS. 152. In this chapter an analytical ke}- leads to a set of con- cise, practical rules, embod^dng most of the ordinary metliods emplo3'ed in solving differential equations ; and the attempt has been made to render these rules so explicit that they may be understood and applied by any one who has mastered the Inte- gral Calculus proper. The key is based upon ' ' Boole's Differential Equations " (London : Macmillan & Co.), to which the student who wishes to become familiar with the theoretical considerations upon which the working rules are based is referred. 153. A differential equation is an expressed relation involv- ing derivatives with or without the primitive variables from which they are derived. For example : {l + x)y + {\-y)x^^ = Q, (1) «'|-«2/ = ^'+l. (2) D^z-aWy^z = 0, (4) are differential equations. The order of a differential equation is the same as that of the derivative of highest order which appears in the equation. Equations (1) and (2) are of the first order; (3) and (4) of the second order. The degree of a differential equation is the same as the power Chap. XV.] DIFFEEEISTTIAL EQUATIONS. KEY. 159 to which the derivative of highest order in the equation is raised, that derivative being supposed to enter into the equation in a rational form. Equations (1), (2), (3), and (4) are all of the first degree. A differential equation is linear when it would be of the first degree if the dependent variable and all its derivatives wei'e regarded as unknown quantities. Equations (2), (3), and (4) are linear. The equation not containing differentials or derivatives, and expressing the most general relation between the primitive vari- ables consistent with the given differential equation, is called its general solution or complete primitive. A general solution will always contain arbitrary constants or arbitrary func- tions. A singular solution of a differential equation is a relation be- tween the primitive variables which satisfies the differential equation b}^ means of the values which it gives to the deriva- tives, but which cannot be obtained from the complete primitive by giving particular values to the arbitrary constants. 154. We shall illustrate the use of the ke}' by solving equa- tions (1), (2), (3), and (4) of Art. 157 b}' its aid. (1) (l+x)y-\-(l-y)x^=0, or (l-hx)yclx-\-(l-y)xdy=0. Beginning at the beginning of the key, we see that we have a single equation, and hence look under I., p. 163 ; it involves ordinary derivatives: we are then directed to II., p. 163; it contains two variables : we go to III., p. 163 ; it is of the first order, IV., p. 163, and of the first degree, V., p. 163. It is reducible to the form — ! — dx -\ dy = 0, X y which comes under Xdx -\- Ydy = 0. 160 INTEGRAL CALCULUS. [Art. 154. Hence we turn to (1), p. 166, and there find the specific direc- tions for its solution. Integrating each term separately, we get '\ogx + x + \ogy-y = c, or log{xy) -^ x — y = c, the required primitive equation, (2) x-£-ay = x-hl. Beginning again at the beginning of the key, we are directed through I., 11., III., IV., to V., p. 163. Looking under V., we see that it will come under either the third or the fourth head. Let us try the fourth; we are referred to (4), p. 167, for specific directions. Obeying instructions, the work is as follows : dx -ay = = 0, xdy — aydx = = 0, dy_ adx :0, y ' X lO; gy-a ;logfl; = :C,. ^"4- = c; = 0, y = Cx% dy dx = aCxf X-l _,_ j^< ^dG dx' (1) Substitute in the given equation rlC aCx" + a;"+i— — aCx" = x +1, dx dx dC~^^dx = 0, fy J 1 I _1_ __ Ql (a-l)cc«-^ ax" Chap. XV.] DIFFEEEISTTIAL EQUATIONS. KEY. 161 Substitute this value for C iu (1), and we get ^a a — 1 the requu-ed primitive. (3) |^+H;^*=o. dxr clx Beginning at the beginning of the key, we are directed tlirough I., II., VII., to (20), p. 171, for our specific instruc- tions. Obej'ing these, our work is as follows : y = Ce"^, dy = mCe'"'° clx, Substitute in the equation, and or m^ + 2 m = ; m = or —2. y = C+C'e-'=' is the solution required. (i) JD,'z-a-Dfz = 0. Beginning at the beginning of the kej', we are directed through I. and IX. to (43), p. 179, for our specific instruc- tions. Obeying these, our work is as follows : dif — a-dx^ = 0, dy —adx =0, (1) dy -f- adx = 0, (2) djxly — ahlqdx = 0. (3) Combining (1) and (3), we get dpdy — adqdy = 0, or dp — adq = 0. (4) 162 INTEGRAL CALCULUS. [Art. 154. (1) gives y — ax = a. (4) gives p — aq = ^. (2) and (3) give us, in the same way, y + ax — tti, and our two first integrals are p-aq=f^{y-ax), (5) p + aq=f^{y + ax), (6) /i and/2 denoting arbitrar}- functions. Determining p and g, from (5) and (6) , i> = i [/2(2/ + «aj) +/i(2/ - "a') ] ' ^ = --[/2(2/ + aa;) -/i(?/ - ao;)] ; dz=^^\_f2{y+ax) +/i(y-ax)] dcc+— [/2(2/+acc) +/i(2/-aa;)]d?/ f\{y + ax) ((^?/ + adx) —f\{y — ax) (dy — adx) "^ 2a Hence, z = F(y + aa;) + i^i (2/ — ax) , where F and F^ denote arbitrary functions obtained b}' integrat- ing /i and /2, which are arbitrary. ——* Page Single equation . I. 163 S^'stem of simultaneous equations .... VIII. 166 I. luA^olving ordiuaiy derivatives II. 163 Involving partial derivatives IX. 166 II. Containing two variables III. 163 Containing three variables and of first degree. General form, Pdx + Qdy + Bdz = . . . (34) 1 75 Containing more than three variables and of the first degree. General form, Pdxi -^Qdx2 + Bdxs-h =0 (35)176 III. Of first order IV. 163 Not of first order VII. 165 IV. Of first degree. General form, 3Idx+]Sfdy=0, V. 163 Not of first degree VI. 164 V. Of first degi-ee. General form, Mdx-{-Ndy ^ ll) = 0. Of or reducible to the form Xdx + Ydy = 0, where X is a function of x alone, and Y is / a function of 2/ alone * (1) 166 M and ^ homogeneous functions of x and y of the same degree (2) 166 Of the form {ax-\-by-\-c)dx -\-(a'x-^b'y-{-c')dy = (3) 167 Linear. General form ~+Xiy=X2, where X^ and Xg are functions of a; alone * .... (4) 167 * Of course, X, X^, X,, and Fmay be constants. 163 164 INTEGEAL CALCULUS. dv ^*^* Of the form -^ +Xi?/=Xo?/", where Xj and X2 dx are functions of x alone* (5)167 Mdx+Ndy an exact differential. Test, DyM = D^N (6) 167 3Ix + Ny = (7)168 Mx-Ny = (8) 168 Of i\iQ form. F^{xy)ydx + F2{xy)xdy = . . (9)168 DyM-D^N ^ fjjjjction of x alone .... (10) 168 D.N- DyM ^ f^j^g^ioj^ of y alone . . . . (11) 168 DyM-D,N f^^^tioj^ of rr^y\ (12) 168 Ny-Mx ' ^ -^^ . ^ ^ xHD^N-D,M) + nNx^ a function of h Mx + Ny X n being any number (13) 168 VI. Not of first degree. Can be solved as an algebraic equation in j3, where p stands for -^ (14) 169 dx Involves onl}^ one of the variables and p, (Li! where p stands for -± . (15) 169 dx Of the form xf^p + yf2p =fsP, where 2^ stands for ^ '. . . . . (16) 169 dx Homogeneous relatively to x and y . . . . (17) 170 Of the form i^((^,i/f) = 0, where ^ and ij/ are dv functions of x, y, and — , such that <^ = a dx il/ = b will lead, on differentiation, to the same differential equation of the second order (18) 170 A singular solution wiU answer . . . .. (19)1 70 * See note, p. 163. KEY. 165 VII. Not of first order. ^^=* Linear, with constant coefficients ; second member zero* (20) 171 Linear, witli constant coefficients ; second member not zero* (21) 171 Of the form (a+6x-)"^+^(a+&a;)"-^?^ + -\- Ly = X, where X is a function of X alonet (22) 172 Either of the primitive variables wanting . (23) 172 cZ"?/ Of the form — ^ = X, X being a function of X alonet (24) 172 Of the form —4 =Y, Y being a function of ?/ alone t (25) 172 Of the form ^=/^^ ....... (26)172 In In— 2 Of the form ^=/^^, • (27)173 dx" -^ dx"- ^ ' Homogeneous on the supposition that x and xj are of the degree 1, -^ of the degree 0, ^ of the degree -1, (28)173 Homogeneous on the supposition that x is of the degree 1 , ?/ of the degree w, -^ of the degree w — 1, ^ of the degree n — 2, , (29) 173 dx- , ,9 Homogeneous relativel}' to ?/, — , -^, . (30) 173 Containing the first power onl}' of the deriva- tive of the highest order (31) 173 Of the form ^, +X$^ + T dx- dx 0, where ~dy_ dx X is a function of x alone and Y a func- tion of y alonet (32) 174 Singular integral will answer (33) 174 * The first member is supposed to contain only those terms involving the depen- dent variable or its derivatives. t See note, p. 163. 166 INTEGEAL CALCULUS. Page VIII. Simultaneous equations of the first order . . (36) 176 Not of the first order (37) 177 IX, All the partial derivatives taken with respect to one of the independent variables . . (38) 177 Of the first order and Linear X. 166 Of the first order and not Linear .... XL 166 Of the second order and containing the deriv- atives of the second order onl}^ in the first degree. General form RD^z -\- SD^DyZ + TDy^z = V, where B, S, T, and V may be functions of x, y, z, D^z, and D^z . . . (43) 179 X. Containing three variables (39) 178 Containing more than three variables . . . (40) 178 XI. Containing three variables (41) 178 Containing more than three variables . . . (42) 179 (1) Of or reducible to the form Xdx + Ycly = 0, where X is a function of x alone and Y" is a function of y alone. Integrate each term separately, and write the sum of their integrals equal to an arbitrary constant. (2) M and N homogeneous functions of x and y of the same degree. Introduce in place of y the new variable v defined by the equation y = vx, and the equation thus obtained can be solved by (1), Or, multiply the equation through by , and its 3fx + ^y first member wiU become an exact differential, and the solution ma}^ be obtained by (6).' KEY. 167 (3) Of the form (ax -j-by + c) dx + (a'x + b'y + c') dy = 0. If a6'— a'6 = 0, the equation ma}' be thrown into the a' form {ax + by -{-c) dx -\- - (ax -\- by -\-c)dy = 0. If now z = ax-\- by be introduced in place of either x or ?/, the re- sulting equation can be solved by (1). If ab'— a'b does not equal zero, the equation can be made homogeneous by assuming x = x'— a, y = y'— ^, and determining a and /3 so that the constant terms in the new values of 31 and iV shall disappear, and it can then be solved by (2) . dv (4) Linear. General form -—-{- Xiy = Xo, where X^ and X2 are functions of x alone. Solve on the supposition that X2 = by ( 1 ) ; and from this solution obtain a A^alue for y involving of course an arbitrary- constant C. Substitute this value of y in the given equation, regarding C as a variable, and there will result a differential equation, involving C and x, whose so- lution by (1) will express C as a function of x. Substitute this value for C in the expression abeady obtained for y, and the result will be the requu-ed solution. dv (5) Of the form -^ -f Xiy = X22/", where X^ and X2 are func- tions of X alone. Divide through by y", and then introduce z = 2/^~" in place of y, and the equation will become linear and may be solved by (4). (6) 3fdx + Ndy an exact differential. Test DyM= D^N. Find I 3Idx^ regarding y as constant, and add an arbi- trary' function of y. Determine this function of y by the fact that the differential of the result just mentioned, taken on the supposition that x is constant, must equal Ndy. Write equal to an arbitrary constant the | Mdx above men- tioned plus the function of y just determined. 168 INTEGRAL CALeULUS. (7) Mx + Ny=0. Multiph' the equation through by , and the Mx — Ny first member will become an exact differential. The solu- tion maj' then be found b}" (6). (8) Mx - Ny = 0. Multiply the equation throuo-h by , and the Mx 4- JSfy first member will become an exact differential. The solu- tion may then be found b}' (6) . (9) Of the form fi(xy)ydx-^f2{xy)xdy = 0: Multiply through by , and the first member Mx — JSfy will become an exact differential. The solution may then be found by (6). (10) DyM—D^N ^ ^ function of x alone. Multipl}^ the equation through b}" e-^ n , and the first member will become an exact difierential. The solution may then be found by (6). (11) D.N-D^M ^ ^ fmiction of y alone. M m^N-DyM Multiply the equation through by e-' m , and the first member will become ai^ exact differential. The solution maj^ then be found by (6). (12) DyM-D,N ^ ^ function of {xy) Ny — Mx i -D^M - D^N- dv Multiply the equation through by &' Ny-Mx where V = xy^ and the first member will become an exact differ- ential. The solution may thus be found by (6). ,^3. x^{D^N-D,M)-\-nNx ^ ^^^^^^.^^^ ^f. ?/ . ^^ ^^.^^^ ^^^. Mx+Ny X number. KEY. ■ 169 r y Multiplj' the equation through hy aj"ey^"''", where "" = ^' and fv = ^'(-Px^^— ^,^^/) + »^^« and the first member will become an exact difterential. The solution may then be found by (6). (14) Can be solved as an algebraic equation in p, where p stands for -^. dx Solve as an algebraic equation in p, and, after trans- posing all the terms to the first member, express the first member as the product of factors of the first order and degree. Write each of these factors separatel}' equal to zero, and find its solution in the form V — c^O \>y (V.). Write the product of the first members of these solutions equal to zero, using the same arbitrary constant in each. (15) Involves onl}' one of the variables and p, where x> stands for '^. dx 'Bj algebraic solution express the variable as an expli- cit function of p, and then differentiate through relativel}^ to the other variable, regarding p as a new variable and remembering that — = — There will result a differen- dy p tial equation of the first order and degi'ee between the sec- ond variable and p which can be solved b}' (1). Elimi- nate p between this solution and the given equation, .and the resulting equation will be the required solution. (16) Of the form xf^p -\-yf2P =fzP^ where p stands for -^. Differentiate the equation relative^ to one of the vari- ables, regarding p as a new variable, and, with the aid of the given equation, eliminate the other original vai'iable. There will result a linear differential equation of the first order between p and the remaining variable, which may be simplified b}^ striking out any factor not containing -L or 170 INTEGEAL CALCULUS. -^, and can be solved b^- (4) . Eliminate j? between this dy solution and the given equation, and the result will be the required solution. (17) Homogeneous relatively to x and y. Let y = vx, and solve algebraically relatively to p or u, p standing for -^. The result will be of the form p> —fvi or V = Fp. If p =fv, - =fv, -^ =fv, X- + V =/., y an equation that can be solved by (1 ) . If ?; = Fp^ - — Fp^ y = xFp, an equation that can be solved by (16). (18) Of the form i^(<^,i/') = 0, where ^ and i/^ are functions of ic, y and — , such that (^ = a and i/^ = 6 will lead, on differ- CtQO . entiation, to the same differential equations of the second order. , ay Eliminate -f- between j2 from the given equation, writing them separately equal to zero, and eliminating p and q between them and the given equation. (42) Of the first order and not linear, containing more than three variables. F{xi^x.2, , a;„, z^pi^p^-t >i>«) = 0, where Pi = Dx^z, 2h = Dx^z, Form the linear partial differential equation %i\^{DxiF + PiD^F)Dp^^ - Dp.F{Dx,^ + PiD^^)'\ = 0, where is an unknown function of (.x'l, , .t,i, Pi, iP„), and where % means the sum of all the terms of the given form that can be obtained by giving i successively the values 1, 2, 3, , n. Form, by (40), its auxiliary sj'stem of ordinary differen- tial equations, and from them get, by (36), n — 1 uite- grals, $1 = «!, o = Oo, , $„_! = a„_i. By these equations and tlie given equation express pi, p2-, , Pn in terms of the original variables, and substitute their values in the equation clz —jhdxi -{-jhdx.^ + +2J„da;,j. Integrate this by (35), and the result will be the required complete primi- tive. (43) Of the second order and containing the derivatives of the second order onl}' in the first degree. General form, BD^-z + SD^D^z + TZ>/z = F, where i?, S, T, and Fmay be functions of a% y, 2;, D^z, and DyZ. Call D^z p and DyZ q. Form first the equation Rdf - Sdxcly + Td^ = 0, [ 1 ] and resolve it, supposing the first number not a complete square, into two equations of the form dy — 7?ii dx = , dy — m^ dec = . [2] From the first of these, and fi'om the equation Rdpdy + Tdqdx — Vdxdy = 0, [3] 180 INTEGRAL CALCULUS. combined if needful witli the equation dz = 2)c?a; + qdy, seek to obtain two integrals Ui = a, Vj = /3. Proceed- ing in the same way with the second equation of [2], seek two other integrals Ug = ai, Vj = /Si ; then the first in- tegrals of the proposed equation will be 'ih=fiV, ih=f2V2, [4] where /i and/2 denote arbitrary functions. To deduce the final integral, we must either integrate one of these, or, determining from the two p and q in terms of X, y, and z, substitute those values in the equation which will then become mtegrabie. Its solution will give the final integral sought. If the values of nii and wi2 are equal, only one first in- tegral will be obtained, and the final solution must be sought b}' its integration. When it is not possible so to combine the auxiliary equations as to obtain two auxiliary integrals u = a, v =(3, no first integral of the proposed equation exists, and this method of solution fails. Examples. (1) sin,:« cos ydx — coscc sinydy = 0. Ans. cosy = c cos a;. (2) [2 a/ (xy) — ic] dy + ydx = 0. Ans. y = ce~ V^. (3) {2x-y + l)dx + {2y-x-l)dy = 0. Ans. x^ — xy + y^ -\- X — y = c. (4) ^4-ycosx = 5H^. Ans. ?/ = sina; - 1 + ce"""'. dx 2 KEY. 181 (5) (l-af)^-xy = axyK Ans. y = lc^{l -x-)-ay\ dx xdy — ydx „ . „ ^ + 2/" , 4-„,,-iy _ (6) xdx + 2/d2/ + i;, ,. = 0. ylns. —2— + tan - _ a^ + 2/^ ' 2 w^=4M-(l)- ^47is. y = ex + c — (?. Singular solution, y = ^^^— 4 (8) /'^Y_£! = 0. Ans. (y-a\ogx-c)(y+alogx-c) = 0. \dxj X- ^ ^ doj* da^^ dx^ dx -^ Ans. ?/ = (co + Cia; + C2a3- + C8a^)€^ (10) pi -2k'M + ^-22/ = e\ Ans. y = (ci+C2o;)e^^+ — ^- dx- dx K'^ — ^) (11) ^-|_i^==o. ^ws. 2/ = c log a; + c'. die- a; dx (12) a;2^' + a;^^+(2x-?/-l)^+?/2 = 0. Find a first integi-al. da;3 dx" dx -.o . .7,. dxr dx (13) _^ + _^ + -^=0. Ans. {x-a){y-b){z-c)=C. x—a y — h z—G (14) (y-{-z)dx-i-dy + dz = 0. Ans. e'{y + z) = c. ' (15) ^' + 4a; + ^=0, ^i^ + 3?/-a^ = 0. d< 4 ai Ans. x = ce2_^, 2/ = (c^+ ^j) e 2. (16)^4-771^0^=0, ^-m2a; = 0. Ans. ^ ' dt' df riT^ Dz = — ^^*«- e'^a; + 2/ + 2) = <^y. ^ ^ ^ a; + 2; 182 INTEGEAL CALCULUS. (18) xzD^z-\-yzDyZ = xy. Ans. z^ = xy -\- 0 1 A» li r 1 1 /, — = vers ^x. JL-'-j. V CI O iO >— — ^{2x-x^) The second, fifth, and seventh in the second group can be written in the more convenient forms, 66 DIFFERENTIAL CALCULUS. [Art. 75. log a y^sina;= — cosa;; f^tsLUX— — logcosaj. 75. When the expression to be integrated does not come under an}^ of the forms in the preceding list, it can often he iwepared for integration by a suitable change of variable, the new variable, of course, being a function of the old. This method is called integration by substitution, and is based upon a formula easily deduced from D^(Fy)=DyFy .D^y ; which gives immediately Fy=A{D,Fy.D^y). Let u=D,Fy, then Fy=fy^h and we have fyU=f^{uD^y); or, interchanging a; and y, fu=^f^{uD,x). [1] For example, required f{a + 6x)". Let ' z = a + bx, and then X(a + bxy =/,z- =f,{z^ . D,x) , by [1] ; , , z a but x= , b b B.x = - ; hence /,(a + bxY = {f^z'^=l ^"^ b bn + 1 Chap. V.] INTEGRATION. Substituting for z its value, we have _1 (a + 5a;)»+^ 67 Jl{a + hx) b 71+1 Example. Find X a + bx Ans. -log (a + bx) . 76. Iffx represents a function that can be integrated, /( a + bx) can always be integi'ated ; for, if then and Find (1) f^sinax. (2) f^cosax. (3) y^tanaa". (4) 7^ ctn ace. 77. Required/^ L z= a + bx, D,x = - fj{a + 6a;) =fjz =fJzD,x = ^^fjz. Examples. Ans. cosaa;. a Ans. —sin ace. a 1 1 yj {a- — X-) a i^-m Let then X Z = -, a x = az, D^x = a, 68 DrETEEENTIAl. CAliCULTIS. 1/. a L = i/. a-'VCl-^') ci'^il-z") [Art. 78. 1 Examples. = sin~^2 = sin~^ -. Find (^\ r 1 X Ans. -tan~^-' ^^^ -Sr+ct-^ (^\ r . X Ans. vers ^-' C6 ^"^ -^VC^a^-^O 78. Required f^—-—, -• V(.^" + «-) ■ Let z = x + ■sj(x^ + a^) ; then z — x = ^{a? + a^) , z^ — 2 zx -i- (kP = x^ -\- a^, 2zx = z- — a^, 0. = ^'-^', 2z ' 2;2 _ 0,2 z'^ + ci? 2z 2z D^x z- + or 2 2- /. =X^,=/.-^^^^^^ ^ ( .^•^ + a-) "^ 2;- + a- 2^ + a^ =/. 2z z^-\-a? ' ~\ , ^ /-5 •„, 72 "ir:^ =•/-- = log2 = log(;« + ^xr + a-) . z^ + ct^ 2 2;^ Example. FindX sji.^-o?) Ans. log {x + Va/*^ — a-) , Chap. V.] INTEGRATION. 69 79. When the expression to he integrated can he factored, the required integi'al can often be obtained by the use of a formula deduced from D^{iiv) = uD^v + vD^u, which gives uv = f^ uD^ v -\-f^ vD^ u or f^uD^v = uv —f^vD^u. [1] This method is called integrating hy parts. (a) For example, required f\ogx. \ogx can be regarded as the product of logic by 1. Call log a; = u and 1 = D^v, then D^u = -, X v = x; and we have ^logcc =/^l logo; =f^uD^v = uv —f^vD^u = a; log a; —fx- = ^loga; — x. X Example. Find y^O/* log cc. Suggestion : Let loga; = u and x = D^v. Ans. - x^ ( loffoj 80. Required fsin^x. Let u = sin a? and D^v = sin a;, then D^it— cos X, v= — cosa;, 7^sin^a;= — sina^cosa; +/^cos^a; ; 70 DIFFEBENTIAIi CALCUI.US. [Art. 81. but cos^x = 1 — srn^x, so Xceos^x=f^l—f^sm-x = x—f^sm^x and /x sin- x = x — sin x cos x — f^ sin^ a;. 2Xsin-a' = x — sin .'c cos a?. f^ sin^ x= ^(x — sin x cos x) . Examples. (1) FindXcos^a;. Ans. -(a; + sina;cos.T). (2) /j-sinxcoscc. Ans. -. 81. Very often both metliods described above are required in the same integration. (a) Required f^sm~'^x. Let sin~\T = ?/, then x= sin?/ ; DyX= cosy, f,sm-^x =f^y =f^ycosy. Let u = y and DyV = cosy ; then A« = l, V = sin?/, and /j, ?/ cos 2/ = ?/ sin ?/ — /j, sin ?/ = ?/ sin ?/ -|- cos ?/ = a; sin" ^T + ^y ( 1 — .1'-) . Any inverse or anti-function can be integrated by this method if the direct function is integi-able. (6) Thus, fJ-'x=f^y=f^yDJy=:yfy-fJy where y=f-^x. Chap. V.] INTEGRATION. 71 Examples. (1) Find/a.cos~^a;. Ans. a;cos~^cc — ^(1 — x^). (2) f^iari~^x. A71S. a;tan~^a; — -log(l + a^). (3) yivers~^£c. Ans. {x—l)yevs~^x-{-^(2x—x^). 82. Sometimes an algebraic transformation, either alone or in combination tvith the preceding methods, is useful. 1 (a) Required/^ af— a- x^— a' 2 a \x — a x + aj and, by Art. 75 (Ex.), A i = ^ Dog(^ - a) - log(.« + a) ] = — log'^^. XT— a' 2 a 2a x-i-a (b) Required/, \(:^^. \\i-xj V(i-^') VCi-^") vci-^-" /x 1 . — oin -*^o' fx — — ^ 5- can be readily obtained by substituting ?/ = (1 — a^, ^(1 — ar) and is — ^/(l — a.*^) ; hence / /('l±|') = sin-^o; _ ^(1 _ ^2) _ (c) Required f^-^/ (a^ — x^) . V(a'-a^) = V (a^ - x') V («' - ^') V (a' - »^) ' 72 DIFFERENTIAL CALCULUS. [Art. 83. and /,V(a^ - ^'^) = a^f. „ } ,, -/» whence /, V («^ - ic^) = a^ sin"^ ^ -/^ ,, f" — 5-, by Art. 77 ; af by integration by parts, if we let w = V (^^^ ~ ^^) ^iicl Z)^ V = 1. Adding our two equations, we have a and . • . X V (a' - a^) = | Ma^-x" + a^ sin"^ ^] • Examples. Find (1) /.^{x' + a'). (2) /,V(a^-^-a^). ^ws- ^ [a; V (^ - a') - aHog(x + Va;- - a^) ] . Applications. 83. To find the area of a segment of a circle. Let the equation of the circle be ar + 2/" = "^ J and let the required segment be cut off by the double ordinates thi'ough {xQ,y^ and {x,y) . Then the requu-ed area A=2f^y + C. Chap. V.] INTEGE.ATION. 73 From the equation of the circle, hence A = 2/^ V («^ -x-) + C; and therefore, b^- Art. 82 (c) , A = x-sJia' - x") + a' sin"^ | + C. As the area is measiu'ed from the ordinate i/q to the ordinate y^ ^ = when X — Xq ; therefore = Xq V (« — -V) + «" sin~ ^ h C*, C = — .To V (f'' — ^'o') — «" sill ^ — ' and "we have ^ t= aj ^ (a^ — it'-) + a^ sin~-^ ct'o V ("^'^ — ^"o^) -~ «' sin~ ^ — If 0^0 = 0, and the segment begins luith the axis ofY, X A = X'^ (a^ — ar) + «^ sin~^— • If, at the same time, x = a, the segment becomes a semicircle, and A= a ^/(a^ — a^) + a^sin"^- = ~ — ct ^ The area of the whole circle is -a^. 74 DEFFEEENTIAIi CALCULUS. [Art. 84. EXAIVIPLES.. (1) Show that, hi the case of an ellipse, .1=' £:+iC=i, a- h- the area of a segment beginning with any ordinate y^ is X -Jia^ — it"') + a^sin"^^ Xf^JCa^— Xq) -^ a^sin" a That if the segment begins with the minor axis, ^=- ccy(«- — ar) + trsm - a I a That the area of the whole ellipse is -ab. (2) The area of a segment of the hj^^erbola cr b' A = x-sJ(x^ ~ a-) — fr log(cc + Vic^ — «') — Xo V(^'o" — «") + «"log(a;o + Va^o^ — a-). If a'o = a, and the segment begins at the vertex, A = X'sJ{a^ — cr) — a- log (a; + V-»2 — ^^2) + a'logrt. 84. To find the length of any arc of a circle, the coordinates of its extremities being (cccyo) ^^^^ i^iV) • By Art. 52, s=/.V[l+ (A^)']- From the equation of tlie circle, x^ + y-= a% Chap. V.J INTEGRATIOlSr. 75 we have 2x-\-2yD^y = 0, y r f s=f^^=af^ } =asin-^^>a (Art. 77.) When cc = i»07 s = ; hence = a sin~ ^ - + C, C = — asm ^-, a' . /^ . _i^ . i-'^o and s = a sin ^ sin~^ — \ a a Ka;o= 0, and ^7ie arc is measured from the highest point of the X c ircle , s = a sin~ ^ - • ' a If the arc is a quadrant, x = a, • _i/i\ T^a s = asin 1(1) = — , and the whole circumference = 2-a. 85. To find the length of an arc of the parabola y^= 2mx. We have 2yD^y = 2m ; 7-v m Ay = — ; y 76 DIFFEEENTIAL CALCULUS. [Art. 85. 5=/. J 'sl{m? + y'^) = fy ~^(m^ + y'^) DyX by Art. 73 ; s = —fy ^/mF+f=—-ly\/7n^ + y^-+ mHogiy + Vm-+ ?/-)] + C, m 2m by Art. 82, Ex. 1. If the arc is measured from the vertex, s = when y = 0; = (m^ log m) + 0^ 2 m C= mlogm, and s = 'y-^jim' + y^) y + ^ (rri^ -^ y-)' +mlog- m Example. Find the length of the arc of the curve a^= 27 ^/^ included be- tween the origin and the point whose abscissa is 15. Ans. 19. Da*' r BOSTON COLLEGE 'O^ i 3 9031 01548937 ON C0U£GF5!Mr ^60413 DOES NOT CiRCULAi B^^ BOSTON COLLEGE LIBRARY UNIVERSITY HEIGHTS CHESTNUT HILL, MASS. Books may be kept for two weeks unless other- wise specified by the Librarian. Two cents a day is charged for each book kept overtime. If you cannot find what you want, ask the Librarian who will be glad to help you. The borrower is responsible for books drawn in his name and for all accruing fines.