EXAMPLES 
 
 INTEGRAL CALCULUS. 
 
 BY JAMES HANN, 
 
 II 
 
 MATHEMATICAL MASTER OF KING'S COLLEGE SCHOOL, LONDON 
 
 HONORARY MEMBER OF THE PHILOSOPHICAL SOCIETY OF NEWCASTLE-UPON-TYNK, 
 
 JOINT AUTHOR OF MECHANICS FOR PRACTICAL MEN, 
 
 AUTHOR OF THEORY OF BRIDGES, JOINT AUTHOR OF NAUTICAL TABLES, 
 
 AUTHOR OF A TREATISE ON THE STEAM ENGINE, 
 
 AUTHOR OF THEORETICAL AND PRACTICAL MECHANICS, 
 
 AND AUTHOR OF PLANE AND SPHERICAL 
 
 TRIGONOMETRY. 
 
 LONDON: 
 JOHN WEALE, 59, HIGH HOLBORN. 
 
 1850. 
 
Hz 
 
 LONDON : 
 GEORGE WOODFALL AND SON, 
 
 ANGEL COURT, SKINNER STREET. 
 
PREFACE. 
 
 As this work contains a great number of Integrals fully 
 worked out, the Author hopes that it will considerably facili- 
 tate the progress of those who are entering on this branch 
 of study, by showing them almost all the artifices that are 
 used in those branches that come within its scope. 
 
 The works that have been consulted are those of Peacock, 
 Gregory, Hall, De Morgan, Young, and various mathema- 
 tical periodicals ; also the excellent little work on the Cal- 
 culus by Mr. Tate, which, like all the productions of that 
 eminent writer, abounds with useful information, apart from 
 the able manner in which he has treated the first principles. 
 
 Where integration by parts is used, the whole process is 
 put down, but the student should endeavour as soon as pos- 
 sible to acquire the facility of running off the quantities 
 without writing down all the intermediate steps. 
 
 303932 
 
/ 
 
 <a/ i*> ^Ty 
 
 4 * *■* **" 
 
 fat 4* &**>* ^ ei*^ ~ ^^ '*"- 
 
 
 
 
 
 3W /^^c 
 
 
 ^ 
 
 ttdtr+rdu ^h & 
 
 * 
 
 €» 
 
 /- isr 
 
 -+ 
 
 6 
 
 / &> 
 
EXAMPLES 
 
 INTEGRAL CALCULUS. 
 
 CHAPTER I. 
 
 ELEMENTARY INTEGRALS TO BE COMMITTED TO MEMORY. 
 
 9 J a + bx* s/ai \ V J J <*+# 
 
 = - tan l -. 
 a a 
 
 dx .__»■# 
 
 = sm~* -. 
 
 •/ va — x- 
 
 — d# , x 
 
 cos ~ 1 -. 
 
 6.) / = vers" 1 -. 
 
 •/ v 2 ax—x l a 
 
 f* d x 
 
 7.) / — = log (#±a+ Jw % ±$ ax. 
 
 J vV±fl ax 
 
 n ^ P dx 1 .# 
 
 8.) 1—7; -=-sec" 1 -. 
 
 »' a; sr x'—a* " a 
 
EXAMPLES ON THE 
 dx 1 , X 
 
 ( io.) /Cfi^ i= iio g ^y 
 
 v V # 2 — a~ S 5 \x + aJ 
 
 (11.) a*dx = , . 
 
 v J J log a 
 
 e™dx = -—. 
 
 Examples. 
 
 (2.) / — T7 — ■ — ? — r. Let a? = -, then the integral is re- 
 V x l {a + bx) z 6 
 
 , p zdz z b . , , N 
 
 duced to / —j = h — log (az + b) 
 
 J az + b a (r ov 
 
 1 b. (a + bx\ 
 = + -log(— v — - . 
 
 , rt v p{adx bdx cdx edx] . b 
 
 c e 
 
 (4.) f(l+ai i )(l+w)ada^ / (l+x + aP+x^xdx 
 
BAL CALCULUS. 
 
 r(l+xf(l—x)dx />l+x~x 2 —x\l.v P( 1 1 
 
 (o V~ ~* = J ? "VK^+x 
 
 \ 7 , a; 2 1 a- log ar — 2 ar— a? 3 — 2 
 
 — 1 — x )d%=: log x—x = 
 
 / ° 2 x 
 
 2x 
 
 /x A d x f* f 1 1 . a; 3 
 
 du= ij — . Leta? + 2=£; dx—dz 
 
 {z—Zfdz __ (z'—Gz^ + lZz^dz 
 ■ (~_6+i?--,W .-. u--£-6*+l».log*+? 
 
 \ Z Z~ J A Z 
 
 _ y»-12g» + g.log(3 g4 )+ 16 
 
 _ (x + 2) 3 -- 12 (a* + 2) 2 + (a? + 2) log Qr - 2) 84 + 10 
 ■* 8 (* + 8) 
 
 = j - 7T + log <V ?+]";. 
 
 /5 cZ a? 1 
 
 -—j — V-> Let tf = -, then the integral is 
 2 a: 4 + o ar z 
 
 /— 5 z- 
 ~2?1 
 
 . ■ — 5 £ - <fo 
 reduced to / ■ t , r 
 
 - o cr 
 
 1 /• L-/* / - 10 d 
 
 ■""J,/ 8+s2 *" ~ ;7 V"*~87+»J 
 
 
EXAMPLES ON THE 
 
 •10 dz r odz^ 
 
 /VlO dz _bdz\ 
 
 j \ 3 * 3**4-2 ~y 
 
 -K«VH-(V£W> 
 
 r oc^J xdx p ( ,- 1 \ 
 
 now— = = / (»— v«+l - d# 
 
 ™V 1 + n/a' */ \ 1+-/*/ 
 
 x 1 %sA f dx 
 
 //J Sfi 
 = = (by substituting *J x =z) 
 1 + \/w 
 
 *M =*/("-&)• 
 
 = 2s— log (1 + zf = 2nAb— log(l + ^a?) 2 ; 
 
 3 
 
 .-. the integral is = ^r — -^-+a?— 2(a?)* + log(l + V ) 2 
 
 (11.) f x y x ' dx . Let ^x = z, then tf=c 2 
 V 1 + x 
 
 dx=% z dz; 
 
 r x 2 V x . dx r z> . , ^ rz Q dz 
 
 
 +-£ — + 2 * — S tan- 1 * 
 
 o 
 
 2 s * v i . « 1 ^w i 1 ] 
 
 = - xi — - # 5 + 2 #* — 2 tan"" x* 
 5 3 
 
INTEGBAJ CALCULUS. 
 
 = -7=. tan _, J - — — \ fan useful integral) 
 
 r dx 1 /• da; __ >2 /» d(%cx + b) 
 
 J a + bx-\-cx~~~ cj tf a~ J (Zcx + bf+ikac— 
 X" +-x 4- - v ' v 
 
 = —j ; tan-M . ■- J. 
 
 yy&ac — b* \\/ ±ac—b~' 
 
 Examples for 1 } ractice. 
 
 (a+bx) 2 
 
 ^fxia + bx^dx ^^^ 
 
 r(a -f bx 2 + car 3 ) (3 car -f "2bx)dx _(a + bx' 1 + cx ) Y 
 
 4 . W(* + ^a) (2 y/* + !)}<** _ ( A . + ^Jt. 
 */ sj so 
 
 no Satf + W 
 5.) / (a + bx)xdx = 3 
 
 0.) r(>/Tf±+»)*» =(a , +N /5rn). 
 
 (a + bx*) 2 .%xbdx = ±-Zt 9 ~. 
 
► EXAMPLES ON THE 
 
 /AX r Sdx s 
 
 J */ x l ~\- x i 
 
 (3# + 2)cZo? 
 
 . ar -f a?* 
 
 -f- * 
 
 i i >t r (px + %)dx 
 
 li#) / / TT = 2VV + aft 
 
 18 A ,^ =-rlog\/a+&a?. 
 
 ■ J a + lx b ° 
 
 . r(2x + l)dx 
 
 •' / A /-o , — - =2^ 3 +m. 
 
 ^ var-f^-fi " 
 
 / 3 SC 5 \ 
 
 l6 A* rd X '+Td X ) dX >/a + bx* + cx h + dx * 
 Va+ bx 2 + ca? 3 + d# 4 ~~ 2d 
 
 ' •/ (a 9 + # 
 
 (a? + t/)i S 
 
 w— 1 
 
 17.) / V ^ n J _ g.(x n + x n - 1 ) g 
 
 L9.) / \ ^2c/ ' __ n.(a + bx + cx 2 )~ . 
 / . i . i\*"~ (n— ro).8c 
 
(I.) To integrate du = 
 
 INTEGRAL CALCULI 
 
 CHAPTER II. 
 
 RATIONAL FRACTIONS. 
 
 xdx 
 
 (*+8)(* + 3) B ' 
 
 _ t x A B P 
 
 Let t — rsr/ — r^V2 = ; — r^ + 7 — r-5% + 
 
 (« + 2) (* + 3)* ~~ (* + I)* (a? + 3) T (a? + 2)' 
 .-. a? = A (a? + 2) + B (a? + 3) (a? + 2) + P (x + 3) 2 , 
 Let a; = — 3, 
 
 .-. — 3 = A (2 — 3) = — A, .-. A = 3, 
 x — 3 (a? + 2) = B (x + 3) (x + 2) + P (a> + 3) 2 
 — 2 (x -f 3) = B (x + 3) (a? + 2) + P (a 4- 3) 2 
 .-. — 2 = B (a? 4- 2) 4- P + 3). 
 Let a? = — 3, 
 
 — 2 = B (2 - 3) = - B, .-. B = 2. 
 Let x = — 2, 
 
 — 2 == P (3 — 2) = P, .*. P = — 2, 
 x 3 2 2 
 
 J ' (* 4- 2)(* 4- 3) 2 {x 4- 3) 2 T (a> + 8) (a; 4- 2)' 
 
 /a?<fo 
 (* + 2)(* + 3)* 
 
 3 . /a? 4- IV 
 
EXAMPLES ON THE 
 
 (2.) To find C t *~f* —r - . Let 
 
 K } J (a? + 2) 2 (a? + 4) 2 
 
 a? A B C D 
 
 r* + t-t-t, + rrr-aa +" 
 
 (a* + 4) 2 (a? + 2) 2 " (a + 4) 2 T (a> + 4) n (a? + 2) 2 ' a; + 2 
 a? 2 = A (a? + 2) 2 + B (a? + 4) (a? + 2) 2 + C (a? + 4) 2 + 
 D (a? + 2) (a; + 4) 2 . 
 
 Leta? + 2 = 0, .\a?== — 2, ar = 4, and C(a? + 4) 2 = 4C, 
 
 .'•• 4 = 4C, .*. C = 1. 
 Let a? + 4 = 0, .-. a? = — 4, a? 2 = 16, andA(a? + 2) 2 = 4A, 
 .-. A =4, 
 ... ** _ ( a + 4)2 _ 4 (^ + 2) 2 = - 4 {a? 2 + 6a? + 8} 
 = -4{(a7 + 2)(a J + 4)}, 
 = B (a? + 4) (a* + 2) 2 + D (a? -f 2) (x + 4) 2 , 
 .% — 4 = B (a? + 2) + D (a? + 4). 
 Let x = — 2, .*. — 4 == 2D, .-. D = — 2. 
 Let a? =—4, .-. — 4 = — 2B, .-. B = 2. 
 
 The fraction reduced becomes, therefore, 
 
 4 2 12 
 
 ;t 2 + Tz-r-x + 
 
 (x + 4) 2 T (x + 4) T (a? + 2) 2 (a? + 2) ' 
 and its integral is, therefore, 
 *&dx p dx 
 
 y "£dx r dx /a? + 4\ 2 
 
 (x + 4J» V(*+8? + ° g VaTT^/ 
 
 ■» 4 da? 
 
 (a? + 4) 2 
 
 yv + %-=/ 4 "* ( * +4) " =; 
 
• RAL CALCULUS. 
 
 — 4 
 
 (x + 4) 
 In the same way, 
 
 y" dx — 1 
 
 f 4 1 -) 5« + 13 
 
 U + 4 m + 2 J ** + G# + 8 
 
 therefore the complete integral is 
 
 #2 + 6# + 8 8 \x + 2/ 
 
 ttx Ax Bx 
 
 Let , , , , w ., , o, = ^r— T + - 
 
 (a? 2 + l)(ar + 3) a; 2 + 1 x' + 3 
 .-. 3 = A (or + 3) -f B(ar + 1), .\ A + B = 0, 
 
 3A + B = 2, .'. A = I, B = — 1, 
 p %xdx __ // xdx xdx \ 
 
 _ /x T TT 
 
 W «■ = (^ + 1)^ + 4) 
 
 b 3 
 
10 EXAMPLES ON THE 
 
 Let , , „ w „ , iN = — — — - + 
 
 (V + 1) (x 2 + 4) (^ + 1) "*" (a* + 4) 
 x 1 = A (a* + 4) + B (x 2 + 1). 
 
 Let # = v — 1, or a?* =— 1, 
 
 .-. — 1 = 3A, ,\ A = — i, 
 
 o 
 ••• ^ + |0e'- + 4) = B(^ + l), 
 
 = — < 2 tan"* 1 - — tan"" 1 x V 
 
 r x n dx —hr SvPd® 
 
 US ' Jd# + 1) (x 2 + 4) ~ 3 J (x> + 1) (*» + 4) 
 
 -I /" { 4 ^+ 4-(ar + 4)}^ 
 "37 («* + l)(** + 4) 
 
 _ 1 /»/ 4J# «f# \ 
 
 "V \ (a? 2 + 4) ~V + 1 / 
 
 ^i^tan-^-tan- 1 ^) 
 
1 ! 
 
 Lot ( 3 *- + *~ 2 ) 
 
 A B C P 
 
 V' "*" T^TZTTv "*" /-* _ ^ "*" T^T 
 
 (»-l) : ' (*-l)* T (*_i) T (** + ]) 
 3*' ! + a>-2=A(*' ! +l) + B(»- 1)(*«+ 1) + G(*-l)»0> + 1) 
 + P(a--1)\ 
 Let.r=l .-. 2 = 2A, .•. A = 1, 
 
 3rf+#— 3-(^+t)=B(*-l)(a?+l)+C(*— 1J»(*>+1) 
 
 + P(*-1) 3 , 
 .-. (2*+ 8)(«- 1) = B(»- 1)(# 2 + 1) + C(«- lf(»> + 1) 
 + F(a-1)'\ 
 .-. (2# + 8) = B(*» + 1) + C(«-, !)(«• + I) + p(*_ J)». 
 Let x = 1 , 
 
 '.-. 5 = 2B, .-. B = -, 
 
 5ar* - 4.1! - 1 
 2 
 
 (5* + l)(as— 1) 
 
 = C(»» +])(»_!) +P(*-])-» 
 
 - 5 * + * - Cfrf +• 1) + P(» -, 1); if# = l, 
 
 .-. - 3 = 2c, .-. c = - : ; 
 
 5« + 1 3, , 
 £— + 7j,» 2 +l) = P(*-l) 
 
12 EXAMPLES ON THE 
 
 
 30 — 2 /30 
 
 \ P = 
 
 =(¥-0 
 
 (m-iy(* + i)Tj (x-iy+zj (x-iy 
 
 § r dx 3 /» 0^0 
 
 "" ay o»-i) + 2,/ (0 2 + i) 
 
 /<#0 
 (tf' + l) 
 
 1 6 X 3 i / n 
 
 8 1/» S« , 
 
 1 5 13. x/^TT . _, 
 
 2(0- l) 2 2 0-12 ° (#- 1) 
 
 , x , (1 — + 2 ) <#0 (1 — + 2 ) c/0 
 
 (6.) <#W = £ 5— i = = 7- r-- — rjr- 
 
 v y 1 H- + 0* + 0* (1 + 0) (1 + SB 1 ) 
 
 1—0 + 2 A B 
 
 Let — - — x ,, , — =r = T^ r -f- 
 
 (1 + *)(1 + 2 ) (1 +0)^(1 +0 2 )' 
 
 . 1 — + 2 = A (1 + 2 ) -J- B (1 + 0) 
 = — 1, 
 
 then 3 = 2A, ,\ A = -, 
 
 q 
 
 I *** + **- -(1 + 2 ) = B(1 +a?) 
 
 .^±^±} = _(i+f)! =B(1 + , ) , 
 
INTEGRAL CALCULUS. 13 
 
 I + X 
 .V B = - 
 
 9 
 
 '(1 — a? + # 2 ) dx 
 
 /(l — a? + a?') «# 
 1 4- # 4- # 2 +"#* 
 
 :/?* _L__I _L \ x \ 7 
 
 = ^(3 log (I 4-*)-tan- 1 a;- i log (1 4- *)* ) 
 
 1/ (i+«y # _, x 
 
 (7.) Jw = 
 
 Let 
 
 x(l +xf (1 4-^4-^0' 
 1 
 
 x (x 4- 1)' (1 +* + ar i ) 
 
 A B 
 
 = r + /1 , ^» + tt-^-n + 
 
 « (1 + #) 2 (1 + *) l+« + rf^y 
 
 .-. 1=A(1+ a) 3 (1 4- * + # 2 ) + B(l 4- x 4- x')x 
 
 + C(l+*)(l+* + x')x 4- P (1 4- xfx. 
 Let # = 0, .-. A = 1, 
 
 1 -(1 +mf(l +#+**) 
 = B(1 4- * + &)*+ C(l+*)(1 4- » + afy* + P(l + 
 .-. — (3 4- 4# 4- 3^ + ^) = B (1 +m +x°~) 
 + C (1 + *) (1 +#+**) + P(l + »y. 
 
14 EXAMPLES ON THE 
 
 Leta? = — 1, .-. B = — 1, 
 
 .-. ~ (3 + 4a? + 3a? 2 + a? 3 ) + (1 + sb + a? 2 ) = 
 C (1 + x) (1 + a? + x 1 ) + p (l + a?) 2 or 
 — (2 + 3 # 4- 2 a? 2 + a? 3 ) = C (1 + x) (1 + x +x*) + P(l + a?) 2 
 -.(2+a?-j-a? ? ) (1 +«)==C(l+a;) (\q.m + at) 4-P(l+a?) 2 
 — (2 + as + sF) = C (1 + ru + a? 9 ) + P (1 + #) 
 a? = — 1, .-. C== — 2, 
 
 ^ (1 4- x) = P (1 + a?), .-. P = x, 
 
 /dx 
 X(i + tf)»(l + A, + a *) = 
 
 pfdx dx %dx ( a?dfa? \ 
 
 ./ W""(l + a?) 2 ~"(l+a?) + l + « + a?V 
 
 = j^-81og(l+*).f 
 
 1 /> /2a? + 3a? 2 + 4a? 3 a? 2 \ 
 
 2 c / * \ a? 2 + a? :l + # 4 a? 2 + a^+P ) 
 
 1 i \/a?* + a? 3 + a? 4 /» x* 1 dx 1 
 
 1 4- a? T ° (1 + a?) 2 ^ a? 2 + #» + *>« 2 
 
! KGUAL CALCULUS. I g 
 
 y xdx / * Zdx n Jldx 
 
 <jj (^ + J? + 75 -J 4 + 4^ + 4^ — J 3 + (j # + iy 
 
 9 /» <• 2 #4-1 
 
 = 37" t &* + ir let -^T** 
 
 1 + 
 
 .-. dx = 
 
 3 
 2 
 
 2 /» da __2 -y/3 /»rfa 1 x»<fe 
 
 3 J (2*-l) 3 ""3' "T "t/'l+^ ""^/S l/l + ^ 
 
 1 . 1 .20 + 1 
 
 = — 7=. tan"" 1 z = —7= tan"" 1 jz~ ■ 
 
 y2 ^ __ 1 , x/^ ■+• ^ + x* 
 • (1 + #)P (1 + • + rf) "" 1 + • + ° g (I + i) 5 
 
 1 .2* + 1 
 
 rr tan~ x jb— . 
 
 N/3 v/ 3 
 
 Or thus, 
 
 1 
 
 # (1 4~ *)" (1 + # + x') 
 1 + a? + ar — # — a? 2 1 1 
 
 #(1 4-#)' 2 (1 + * + #*) ^(l+«y 0-+*) (l+^ + * a ) 
 
 1 4- # — a? C 1 4- on 4- a? — x — . 
 = a?(l 4- #) 2 " 1(1 + *) (l+a?4-*r) i 
 
16 EXAMPLES ON THE 
 
 1 1 1 # 
 
 + 
 
 #(l+#) (l+x) 2 I+# (l+«+ar) 
 
 1 + x — x 1 I x 
 
 X 
 
 (l+o?) l+x) 2 (l+x) l+x + tf 
 
 1 2 1 X 
 
 ~~ * ~~ i + x "" (i + *)* + (i + + « 2 y 
 
 x~+\ + ° g 10 + 1)4 V (1 + a? +"5 5 )' 
 
 And £'**!>• = /ft±ft=il* 
 ,/(l+*+^) ,/ i(» + i) 2 + |J 
 
 / 1 , /2#+l\ 
 
 = log x /^ + , + l -_tan-(- 7r ) ) 
 
 *\Ao + *o 2 (i + ^"+^) ~~ 
 ^tt + log — Try — Ti^l-TT/ 1 
 
 ^'J& + x l — rf—tf ~l/ & ;i O + l)'0* + 1)0-1)' 
 
 Let- 
 
 *0 + i) 3 2 + 1)0-0 
 
 A + B# C D E P^ 
 
 O^+T) + + 1 ) 2 + 0+ 1 ) (* - !) + «?' 
 
in'ik<,!;ai. « aij ii 17 
 
 .-. 1 = (A + B v) (* — 1) (m + tf & + 
 C(o;-4- 1)(#-1)> + D(#* + 1)(*+ 1)(*-1)^ + 
 E(*> + 1) (* + 1)\? 3 + P(# 2 + 1) (* + 1) 2 (>- 1). 
 
 Let a? = \J — 1 , 
 .-. l = (A4-bn/^)(14-^ : ^M^^-i)x-\ / ^T= 
 (A + B >/~^7) (2 >/"ZTT — 2) = 
 2 A s/~^l — 2 A — 2 n/^Tlb-2B, 
 
 .'. by equating 2 A n/ ~T = 2B >/^7, .% A = B, 
 
 2A4-2B = — 1, .-. A=B= — |, 
 
 .% 1 + j (*+ l) 3 (#- 1) ^ = 1 (# 7 4- 2# u - 2# 4 -.s 3 + 4) 
 ■a - (rf + 1) (a 5 + 2# 4 — jp — 4ar + 4) 
 
 = (#»+ 1) {C(#- 1) *» + D(* + 1) (#-1)4* 
 4- E(* 4- 1) 2 *»+P(* + I) 2 (*- 1)}, 
 
 *'• I (^ + ^ ~~ *f "" 4a;2 + 4 ) = 
 
 C (a - 1) a? 3 4- D (a? 4- 1) (x - 1) # 3 4- E (m 4- I)V 
 f ?<•+■!/ (* — 1). 
 
 Leta;= — 1, .-. - = 2C, .-. C= -, 
 2 4 
 
 .-. j* (a? 5 4- W — # 3 — 4«P 4- 4 - ^ 4 4- **) 
 
18 EXAMPLES ON THE 
 
 4 ** 
 
 = (a? + 1 ) { D (* - 1 ) * + E + * ) ^ + p (* + X ) - 2 ) 1 > 
 
 .-. I(^_4a? + 4) 
 
 = B (a? - 1) a? + E (a? + 1) a? + P (a? + 1) (a - 1). 
 
 9 9 
 
 Leta?=-1, .-. - = 2D, .-. D = -, 
 
 ... 1(^_ 4a? + 4 ) _|(^_^) = _i(7*S-9s* + 8a-8) 
 
 _ *( # + i) (7 a? 3 - 16a? 2 + 16 a? -8) 
 
 8 
 
 = (* + 1) {Ear 3 + P (a - 1)}, 
 ... - l(7a? 3 - 16a? 2 + 16a? -8) = Ea? 3 + P(a?-l). 
 
 Let a? = 1, .-. E = -, 
 
 o 
 ... - i (8a? 3 - 16a? 2 + 16a: - 8) = P (* - 1), 
 
 ... _ (# _i) (^ - a? + 1) = P (a? - 1), 
 .-. p = — (<^_ # + i), 
 
 1 1 1 + x 
 
 x % + x 7 — x l — a/' ~~ 4 1+ x l 
 
 119 1 1 1 & — x + 1 
 
 + 4 ' (1 + *) 2 + 8* 1 +a? + 8 ' m — 1 k^ ' 
 
INTEGRAL CALCULUS. 10 
 
 /(lx _ 1 /» \ + * j 
 
 1 P dx 9 p dx 
 
 ] . /» dx p dx p dx p dx 
 
 + 8 J x— 1 ~ J ~x~ V 7 "",/ tf 7 
 
 I p%xdx 1 /» 'J# 1 p dx 
 
 ~~8j T+lP~lJ ai l + L + ZJ x + lf 
 
 = 5T3 - " - q lo 8 C 1 + *") — J tan * T - 
 
 8 a 1 x 8 5V y 4 ■ ' 4 {as + 1) 
 
 + g{log(« + l) 9 +log(#- l)-log^} 
 _ 2+'Zx — 4x' i -4.x — x 2 
 
 + | {log (* 4- l) 9 + log [m ~ 1) "». log (l + **) - log**} 
 
 1 . 2 — 2# — 5* 2 
 
 — - tan"" 1 # ss - 
 
 4 4# 2 (l+#) 
 
 + l°o \f - — V a — t — T tan" 1 *, 
 
 (0) r(?>x-Z)dx 
 
 54?— 2 A B C 
 
 x' + e ** 4- e r #•+» * + -* 
 
 5« — 2 = A O + -J) U 4- 4) 4- JSx (x + 4) 4- C# (# 4- 2). 
 
20 
 
 EXAMPLES ON THE 
 
 Let x = 
 
 - 2= 8A, .'. A = - ] 
 
 4 
 
 x = — 2 
 
 — 12 = — 4B, .-. B = 3 
 
 x = — 4 
 
 11 
 
 — 22= 8C, .*. C= - 
 
 4 
 
 
 /» (5 # — 2) d# 
 
 \ rdx P dx 11 /» <7# 
 
 = — j log x + — log (a? + 2) - j log (m -V 4) 
 
 _1 (x + 2) 13 
 
 ~~4 g #(tf + 4) 11 ' 
 
 (*>/£&£ 
 
 3# + 1 _A B C 
 
 x :i + %x 2 -\-x"~ x (x + l) 2 # + l' 
 3# + 1 = A(# + I) 2 + Bx + C# (a? + 1). 
 
 Let .i? = 0, 1 = A, 
 
 3x + l~x 2 —%x — l = x(l — x) = Bx-\-Cx(x + l), 
 ,\ 1—0 = B + C(0 + 1). 
 
 Let m = — 1, 2 = B, 
 
 ... _ (i + #) = c(« + l), .-. c = — l, 
 
 /(3'# -f 1) df# __ /• J# /* dx r dx 
 
 x> + 2x 2 + x~J V + J {a + iy J a + 1 
 
TNTEGRAL CALCUT' 2L 
 
 2 
 == ] °% a ' ~ 7+1 ~" l0 ° (* + ] ) 
 
 = log 
 
 (11.) /* ^ 
 
 a? + 1 a? + 1 
 dx 
 
 1 A B 
 
 Let r — — : — r + 
 
 1 + 3a? + 2ar » + l T J2.t? + l' 
 1 = A (2 a? + 1) + B(a? + 1). 
 Let a? = - 1 1 = - A 
 
 — jj 1 = +| B-8, 
 
 J a? /* <Za? /* 2rfa? 
 
 r dx ___ /^tfa? /» 2< 
 
 * V T~4- 3a? + 2a?-"" ~~JaT+l + J 2a?" 
 
 = - log (a? + l) 4- log (2a? 4- 1) = log 
 
 4- i 
 2a; 4- 1 
 
 x 4- 1 
 
 r__jcdx___ 
 K V (a?-2)(a?4-3) 2 
 
 a? ABC 
 
 (a?— 2) (a? 4- 3) a (a? 4~ 3)~ ^ a? 4" 3 a? - 2 ' 
 
 x = A (a? - 2) 4- (a? 4- 3) {B (x _ 2) + C (a? + 3)}. 
 
 3 
 
 ] si m = — 3, .-. — 3 = — 5 A, A = -, 
 
 5a? — 3a?+ 6 . 2 (a? 4- S] 
 
 ""' 5~ " ~~ 5 
 
 ==(* + 8){B(«-8) + C(* + 3)} 
 
 A J-=B(a?-2) + C(*4- 
 
22 EXAMPLES ON THE 
 
 Q 2 
 
 Let x « — a; - = - 5B, .-. B = — — , 
 « - 3; ^ = 5C, ,.c m |, 
 
 y^~2)^+3) 3 
 
 (* + 3) 3 ~~ zSjx + 3 + 25^ - 2 
 
 /(* + 3) 3 ~~ 25Jx + 3 + 25 J 
 3 1 2 , , ^ 2 
 
 _log(* + 3)+--log(*-2) 
 
 5 # + 3 25 ° v ' ' ' 25 
 
 2 , # — 2 
 
 W /E 
 
 25 ° a? -{- 3 5 a + 3 
 (^ + 3# + l)'J* 
 
 W -\- os 1 — 2# 
 
 t a* + 8* + 1 A B C 
 
 Let "3^ 5 S~" = ~ + f + 
 
 & + #* — 3a # a? — 1 # + 2' 
 a?*'+8ff + lasA(«-.l) (a? + 2) + Bx(x + 2) + Car(a?— 1) 
 
 1 
 
 Let # = 
 
 
 
 1 = -2A, ••• A= -2 
 
 a? = 
 
 1 
 
 5= 3B, .-. B= | 
 
 x = 
 
 -2 
 
 -1= 6C, .\ C = -i 
 
 War + 3a? + I) da 
 
 2a; 
 
 \ P dx 5 /* flta 1 r dx 
 
 ~~~%J !v + 3j x— 1 "~ 6 e/ a? + 2 
 
INTEGRAL CM 23 
 
 = - -log * + - 3 log O - 1) - - log (« + 2) 
 
 v *V^ + 1) (x + 2) (*» + 1) 
 
 x _ A B P 
 
 (*-K)(« 4* &)(**+ 1) ~"* + i + fl? + 2 + V + l 
 
 *=rA(*+8) (**+l) +B(a? + 1) (**1)4-P(«+1) (* + 2). 
 Leta?= — 1; —1= 2 A, .-. A = 
 
 2 
 
 o?= — 2; — 2= — 5B, .-. B = + - 
 
 5 
 
 . x 1 + 2*- + x + 2 2 x % + a* 8 + 2* -f 2 
 
 2 
 
 10 
 
 a? + 3 
 
 = P (a? + 1) (x + 2), 
 
 .-. P = • 
 
 10 
 xdx 
 
 /* xdx 
 
 (tf + l)(> + 2)(ar + 1) 
 
 9 ./ + J . 5^7 « + 2 + To7 ?+l 
 
 = ~ Jl°g(* + 1) + jl0g(* + 2) + ^log(«" + 1) 
 + — tan- 1 *, 
 
24 EXAMPLES ON THE 
 
 log y^n. + io tan * 
 
 W/iTO 
 
 + 3 
 
 1 A B M a? + N 
 
 Let ____ = __ + —_ + 
 
 a? 4 + 4a? + 3 (a? + I) 2 a? + 1 * 2 - 2a? + 3 
 
 1 = A (a? 2 - 2a? -f 3) + B(a? + "0 (^ - &# + 3) 
 
 + (M«+N)(« + I) 2 . 
 
 Leta?» — l"i 1 = 6A .-. A=- 
 
 6 — ar g + 2a?— 3 _ (a? 2 — 2 a? - 3) _ (a?+ 1) (a?— 3) 
 
 6 " ~~ 6 6 
 
 = B(a? + 1) (a? 2 — 2a? + 3) + (Ma? + N) (a? + l) 2 , 
 a;- 3 
 
 G 
 
 = B (a? 2 — 2a? + 3) + (Ma? + N) (a? + 1). 
 
 Leta? = ~ 1; o = 6b •• B = n> 
 o 9 
 
 — <dx + 27 — 6a: 2 + 12* — 18 
 54 
 
 = "~ (a * > ~*~ 8) = (M»+N)( a > + l) 
 
INTEGRAL CALCULUS. 
 
 dx 
 
 dx 
 
 • r = 
 
 "J a? 4 + 4 x -f 3 
 
 I p dx \ p dx % \ p t 
 
 $J (a? + 1)"" + Sy .7TT /+ 0y a?-- 8* + 3 
 
 1 /* a?f/*c 
 
 / 1 I \ p dx 
 
 + U ~~9J J a?- — 2ar + 3 
 
 1 ] 1 a? + 1 
 
 " ~ 6 F+l + ° g ^/**-2* + 3 
 
 1 . *— 1 
 
 H tan -1 . 
 
 T 18 ^/2 «/2 
 
 <W/^-(«-^.-')--/*Tr * 
 
26 EXAMPLES OX THE 
 
 dz 
 
 /* z 5 dz rz 2 dz b Paz 2 d. 
 
 az :i + b ~~ J a d l J az :i + 
 
 - b ' z 2 
 
 az 2 dz 
 b 
 
 -S + B I ?«^ + * ) 
 
 1 b /a+bx*\ 
 
 ~~3^F J + 37 2 ° g l x* J 
 
 — tt— — 5n3 • Let # = - , J# = dz 
 
 x(i +»'0 « * 
 
 /d a? r> z 1 dz _ r* z^dz 
 
 y"> z 2 dz p z 2 dz 
 
 (FTT) + J{J + l) 2 
 
 It,, x 1 
 
 = - O l0 S (*' + 1) ~ 
 
 V ar* + 1 a? 3 
 
 A/ a 3 3(ar» + l)' 
 
 The following method of doing the last four integrals is 
 very simple, and can be often used with advantage. 
 
 /» dx 
 xlx~+T) 
 
 _L (1 + x*) - a? 
 
 x(x' + 1)" x(x* + 1) 
 
 ~~ x x :i -f l' 
 
INTKtiRAL CAI.< I' I ; 27 
 
 — — may be found in a similar manner 
 
 X\\ -j- x ) 
 
 1 (1 + of) — at 1 s 3 
 
 a?(l + # 4 ) *(1 + # 4 ) as (1 + # 4 ) 
 
 y" da? , K f X* 
 
 •*• y^(i + **) ~ og v ^n 
 
 /• da? 
 
 a? 4 (0 -j- b&) 
 
 Here 1 1/ a + ^-^ N 
 
 a? 4 (a + 6a? :1 ) a V & (a + &*») / 
 
 aa? 4 a a? (a + 5a; 3 ) 
 
 1 1 a + k 3 - ^ 
 
 a? (a 4- bat*) a x(a-\-bx') 
 
 J_ bx 2 
 
 ax a(a -\- bx*) 
 
 Therefore, the original quantity is reduced to 
 1 I V x 2 
 
 az* d l x a 1 (a -\- 6 or) 
 
 , . . , . 1 b , i a + bx\ 
 
 and its integral is — - — ■.. + n-^.log — ) 
 
 3aar da~ ° \ x i J 
 
 /dx 
 a (1 + *')*' 
 
 1 (1 + ^)_^ 
 
 1 x l 
 
 1 !+•*— #" 1 a? 2 
 
 #(l+#0 "" o?(l +«■'*) " a; «*+l* 
 
 c 3 
 
28 EXAMPLES ON THE 
 
 1 
 
 Therefore 
 
 x(l + *•)■ 
 
 x 2 
 
 x x* + l (# 3 + l) 2 ' 
 and the integral required is therefore 
 3 /~x T ~ 1 
 
 y^ d# 1 /* r d# dx y 
 
 tf^Ti = ^y \tf - i "" # 3 + if 
 
 ^rn -y^ ~ i) (** + x + i)' 
 
 1 A B^ + C 
 
 Let -r-: = i ; + 3 
 
 x' 3 — 1 0—1 # 2 + + 1 
 1 = A(x 2 4- + 1) + B0 + c(# — 1). 
 
 Let ^ + 0+1=0, or x = v "~ 8 "~ 1 
 
 .-. 4 = (2C - 4B) v/^I- 6C, 
 
 2 1 
 
 ,\ C = . B = • 
 
 3' 8 
 
 Let a? = 1, .". 1 =3 A, A=-, 
 
 o 
 
INTEGRAL CALCULUS. 
 
 .-. the integral is reduced to 
 
 1 n ( dx (x + %)dx ) 
 
 3 J U — 1 ~~ x l + x + 1 ) 
 
 - 5 + 2) d* 
 
 =/- 
 
 "m, 
 
 *+-: 
 
 = lo g A/* 2 - 
 
 - + -7= tan —tz 
 4 v"s V 3 
 
 /- 3 _,/2^— 1\ 
 
 = log v/a; 2 + a? + 1 + — = tan ( — - ) • 
 
 •'•yirzT = 3 lo s( a7 - 1 )-3 lo S >/*» + *+ 1 
 
 - — tan->( — -). 
 
 V3 V n/3 / 
 
 And in a similar manner it may be shown that 
 
 f^h=l lo s (• + J ) - I lo * >/*-' + * 
 
 1 /2« + 1\ 
 
 /> d* 1 . /# — 1 n/#- — a; + 1 \ 
 
 *"- 1 == « ° S \*+ 1 vV> * + 1/ 
 
 2</3 l W3 / T \ v/3 /) 
 
30 EXAMPLES ON THE 
 
 But these inverse tangents together are equal to 
 Ax 
 
 tan- 1 ! 
 
 TT " -'(^) 
 
 -mi 
 
 / dx 1 / a? — 1 \/V — ^ 4- i \ 
 
 # 6 - 1 ~" 6 ° g V^ + 1 y'V + a? + 1 / 
 
 •^ v P x""dx 
 
 WJWTW 
 
 ft* (1 + ar) & — ar 3 
 
 (3? + 1)* ~ p 8 + l) 3 
 
 And 
 
 (^ + If (a? + if 
 «j* #(# 2 4- 1) — a? 
 
 (^ + l) 3 ~ (a 3 + I) 9 
 
 # x 
 
 Also 
 
 # 2 4- l («« + iy» 
 
 x* x (x 2 4- 1) — a; 
 
 (1 +«^F (a 2 + If 
 
 x 
 
 (? + i) 2 "* p? 4- l) 3 ' 
 
 " (« 3 +l) 3 * ' (* 3 + If {x 2 + I) 3 
 
 a? 2a; x 
 
 x 2 4- 1 ~~ (a? 3 + l) 3 + (# + l) s5 
 
r, CALCULUS. 
 
 and the integral required is therefore 
 
 log s/FTT + -r^-j - ^j—r^ 
 
 0' rt +l) = ( ar—Sx cos - -fl) (& — &r cos— + lY- 
 
 continued to the factor ( x 1 — %x cos tt-1-1 ) when m 
 
 \ m J 
 
 is an even number. 
 
 This gives 
 (x { + 1) = far — R* cos 2+l) (^ — 2.« cos-^ + l^ 
 
 or (a; 4 + 1) = (.^ - tiV^ + 1) (** + * -v/2 + 1). 
 
 „. W 7T . «7T 1 3?T . It 
 
 Since, 7- = 45 cos -= sin -= —-, cos -7- ad —sin—. 
 4 44 ^/2 4 4 
 
 Assume therefore 
 
 x" A a? + B C a + D 
 
 **+] "~ * 2 — zn/2 + 1 a^+WS+l' 
 
 a# 2 =(a# + b)(# 2 + W2+1) +(ci+i>)(« , -W5+i). 
 if /*•+ ,Whi = ft, == "~*_T~ l (1.) 
 
 Ifar-WS + l =0, x = _z±±_I (2.) 
 
 v9 
 
32 EXAMPLES ON THE 
 
 Making (1) our supposition, we have 
 
 ,s = J^EL- •=-, aa c ^" c + P -(9--a^i), 
 
 — %C*/~\ 2 0^"-! 2C , 2C 
 
 or, - V-l = — — -f — 7^ + -7= 
 
 V2 V2 >/2 ^2 
 
 + 2D -2D n/~- ^^={^^ 9D f v ^-5 +2D, 
 
 ,\ D=0 C = — • 
 
 2V2 
 
 Now making (2) our supposition, we have 
 
 = (by reduction, as in the preceding case,) 
 {^+9b}V— 1 + 8B, ... B = 0, A = ^- 
 
 The quantity under consideration is now reduced to 
 
 1 ( x x 1 
 
 2\/g Xx 1 — x V2 + 1 «*+ * V2 + 1J' 
 
 a? # 
 
 The integrals of 7= and — — now re- 
 
 x 2 —xv-2+ 1 «H«v2 + ] 
 
 main to be found. They can both be included in the general 
 
 x 
 
 case 
 
 / ^±#^2 + 1 \ n/2/ •/ 
 
 1 , A Z+ Sl) d * 
 
INTEGRAL CALCULUS. 33 
 
 - tan- (,v/5). This %^f x ,^\ + l 
 = log \/{x l - x s/% + l) + ton" 1 0»v/2- 1), 
 
 log a A 2 "^ H1 + tan-'(*/2-l) 
 V a' + av/a + l 
 
 + tan- ! («v/2 + l). 
 And these inverse tangents may be reduced thus : 
 
 tan-^v/a-l) + tan" 1 (x>/% + 1) 
 
 / x V2 — 1 + W5 + 1 \ 
 
 \i-(#V2- i)(W2 + i)/ 
 
 = tan" 
 
 = tan x - — —-— — -\ = tan l -rrr — — = tan ,- r> v 
 
 1 — (2 ^— l) 2 (1-ar) (1 — tr) 
 
 /ardx 
 x r +i = 
 
 _L_ . A A--Va+i 1 f^v/ai 
 
 «? + 1 _* + 1 
 
 1 A B„ + C 
 
 _. ________ __. _______ I I 
 
 & + 1 # + 1 ar — x + l' 
 
 c C 
 
34 
 
 EXAMPLES ON THE 
 
 
 .*. 1 = A O 3 — X + 1) 4- Bx + C (x + 1) 
 
 
 # = — 1, ,\ I = 3 A, .-. A=-, 
 
 o 
 
 3 — (g* — a + l) _ a? 2 + # + 2 
 #4-1 ~ a? 4- 1~" 
 
 #* — 4- 2a 4- 2 
 
 1 1 2 — x 
 
 ■ + ■ 
 
 ar 3 4- 1 3 (« 4- 1) a 3 — * 4- J 
 
 ^ • 7 
 
 - dz 7 
 
 » 2 p zdz 
 
 *+\ 
 
 4 4 
 
 4 
 
 1 ,/2«— 1\ , /— 
 
 = —7= tan -1 ( =-? J — log </a? — * + 1, 
 
 y^ da? 1 j . /2a?— 1\ 
 
 ^n = 7 --tan-(- 7r ) + 
 
 log^EEL, 
 
 \/V— -a? 4- 1 
 
INTEGRAL CALCULUS. 
 Also, I x\lx = —, I dx = X, 
 
 therefore the original integral is 
 
 x* 1 . (2*- 1) , . \/JTl 
 
 T — m + -t= tan" 1 4 —-\ + leg / ^ = 
 
 The method of integrating by parts is used to great advan- 
 tage in many integrals, "which is as follows : — 
 
 d(pq) =pdq + qdp, 
 .\ p> dq = d (j)q) — q dp 
 
 fpdq=zpf — fqdp (1); 
 
 or by using differential coefficients 
 
 /'fs-»«-/«2 » 
 
 In the following examples we shall sometimes use (1) and 
 sometimes (2). 
 
 When integrals are of the form of fat*- 1 (a + bx y ')l d x, 
 
 they can be rationalized by assuming a -f bx n = *; 7 when 
 
 m vi p . . _ . ?>* . . 
 
 — or h W wa integer. If — be a fraction assume a -f- £.6*" 
 
 ?t n ? ° ft 
 
16 EXAMPLES ON THE 
 
 CHAPTEK III. 
 
 (1.) fxdxs/a + x — fdx(a + x — a) \/a + x 
 ~ f dx (a -\- xf — a f dx sj a + x 
 
 
 , x du 1 \/# + « — \/# 
 (2.) -— = -7 7= = ■* z— ' 
 
 393 
 
 7 771 77» 
 
 (B.) ^- = (a - x)(b ~ xf = [b -- x + a --b) (b - x)~ n 
 
 (X X 
 
 
 - + 2 _ + j 
 
 m + 2?i ??i -}- ?i) 
 
 
INTEGRAL CALCULUS. 37 
 
 /i n d U I a \ / 
 
 (4.) ^ = (ar + a) s/ x- + 4a 
 
 = (ar + 4a — 3a) s/x' + 4a 
 
 = (** + 4a) a — 3a >/^ + 4a. 
 To integrate (or + 4a) 2 (integrating by parts). 
 
 Let » = (*» + 4a) 2 ^ = 1 
 
 a# 
 
 rfj^ar 8 + 4a) 2 = *(4> + 4a) 2 — 3 /Va** (rf + 4 a) 
 
 =s#(^ 4- 4a) 2 - 3 /dx(x 2 +Aaf + 12 a /rf* (#■ + 4a) 4 , 
 
 .:Jdx{ar + 4a) 2 = |(*°- + 4 a) 5 + 3a /^(^ + 4 a/, 
 
 .-. * = ?(** + 4a)*. 
 
 4 
 
 (5.)— = Leta + # = * 2 
 
 K ' dx xy/x -f a 
 
 (hi du das 1 2 
 
 2* = - 
 
 a 7 * dx dz (z l — a) ; 
 
 \A1~-\A z + y/a) 
 
 I , z — \/a 1 
 
 M = -7= log ~= = -T^ 10 ^ " 
 
 \/a * + y/a y/« (* + v /a)~ 
 
38 EXAMPLES ON THE 
 = — log-— ^ _ 
 
 sj a s/ a -f x + \/a 
 
 ("v ~r~ = • -Let a -f ox* = ** 
 
 dx x 
 
 z 2 — a\i d# 1 2# 
 
 
 c?ia du dx $ z %z 
 
 dz~~ dx' dz ""(^rr^J < 8**(^ — <t)S 
 
 2 s 3 
 
 3(s*-a) 3 L +, * 8 -aJ 
 ~3l 1+ 2 U-v/S * +v /J> 
 
 3 3 2 ^ _ ^/ a 
 
 2# 2 \/a i z + v« 
 
 = IT" -3-^7^; 
 
 __ 2 \/fl + bx* 2 \/» i v « + bx* + v^^ 
 
 3 a - log T^r- 
 
 __ 2 >/« + bx s 2 \/a \/6 -f ax~* + \/ax~' 3 
 
 - s 3- log -71 • 
 
INTEGKAL CALCULUS. 39 
 
 (1 u 1 + # 2 n 2 
 
 ( •' dx (] —V) sj\ + x x (\ 
 
 (i-Jvi*' 
 
 i + » 
 
 g-*)Vg-*)^' ■ 
 
 = — , ( " — •» = «), 
 •'•" = "7I l0g {7^vi} 
 
 = -Llo g V^Ta+y/8 ) 
 
 v/a I z S 
 
 x/a i - •» 
 
 , ■. f dx _ p dx 
 
 (8 'v v/zr^TT ~ J VW+W+% 
 
 -kg(v/(* + ly+f +*+ J} 
 
 = log {2v/«" + * + 1 + »« + 1} + C. 
 
 
40 EXAMPLES ON THE 
 
 1 
 
 . , *"~2 . - 2x- 1 
 = sin -1 = sm L . 
 
 2 
 
 (io.) /• „ _^ *", • Wi-i- **, 
 
 # = 1— £ 2 J# = — %zdz 
 
 i + # = 2 — s' 
 
 ,2 
 
 
 ydx _ r dz 
 
 (1 + a?) 'y/l—x " ^" j 
 
 " ~2v/2(y</2+* v \/2-*i 
 
 1 , ^2 — * 1 . 2 + ^ — 2^2* 
 
 = 7^ l0g 7M : I = 72 l0g — 2^7- 
 
 1 , 3 — a; — 2 \/2 \/l — a^ 
 "" V2 ° g 1+* 
 
 , X r dX T . 1 7 ^ 
 
 (11.) / J— , 2 . Leta = - ^£== - 
 
 /dz p dz 
 
 = — log {s/ I +Z + Z i + z + J} 
 " - log 1 ~ x ) 
 
INTEGRAL CALCULUS. 41 
 
 i x 
 
 = log 
 
 % \/l + x +& + 2 + x 
 
 _ x (2 + x — 2y/l + x + ar) 
 
 2 + # — 2 \/l + #+a^ 
 
 ~^ -3* 
 
 
 rf(^) 
 
 
 rf? V * _1 dx 
 
 
 v y tf x */<£—& Wa 4 - (or) 2 
 
 
 1 . _,*■ 
 
 
 ma) '•- x 
 
 1 
 
 rttf V(l — a?) (» + a) V2- 
 
 — a? — x l 
 
 2 
 
 
 V9 — (2* + I)"' 
 
 
 . , 2#+ 1 
 .*. u = sm ■ — . 
 
 o 
 
 
 rn) rfM - 
 
 
 * « V (^ - a 2 ) (6 2 - x' 1 ) 
 
 
 x 1 
 
 
 *J x l — a 1 V ' \r — a 2 — (x* — a 2 ) 
 .: u = sin- 1 A A ~ < 
 
 (15.) * 
 
 <** * V« s — <r V6» — ar 
 
42 
 
 
 EXAMPLES ON THE 
 
 
 
 1 
 
 *t X = - 
 
 z 
 
 dx 1 
 
 dz " z 2 
 
 
 
 du du 
 
 dx z* 
 
 
 1 
 
 dz dx 
 
 dz Vl — a 2 z 2 S/6V - 
 
 - 1 
 
 #' 
 
 
 a 2 * 1 
 
 
 
 n/1 
 
 « • ~\/ l — T^ — C 1 — a *") 
 
 .•.?,« = -—• sin- 1 
 a6 
 
 1 — g 2 z 2 ]l 
 
 1 . h A ./JF+ a 2 
 
 : -- sin-* -A / ^ - 
 
 00-) £- 
 
 rf « (l+^v/l-* 3 
 
 (1 + «)a/2(1 + *) - (1 +*)* 
 .-. M = — a/2 (1 + *)-'— 1 
 
 --V i+5- ls= ~ / V m 
 
 m £- 
 
 # 
 
 (i+**)\/i— a 4 
 
 0(1 + a 8 ; 
 
 - (1 + ^ 2 )v/2 (1 + 0*0 - (1 + x 2 ) 2 s/% (1 + x 2 )- 1 - l' 
 
 •*• » « - ^\/« ( L + ^) _1 ~ 1 = - i V ] r 
 
 + x 2 
 
INTEGRAL CALCULUS. 
 
 /1Q\ (}U (^ + \/l+« 3 )' 
 
 ( l8 iz == 
 
 dz ~ ^/l + x 1 
 
 =(-== + l) (* + y/r+*y~\ 
 
 (i9.) *f- ; 
 
 <*# (1 + x) sj\ + - ? 
 
 Let 1 + a? = *, *' = (* — l) 2 , 
 o*w o*m da; 1 
 
 z dx dz z \/z (a 
 
 r-l)« 
 
 zs /?>z-z l - 1 
 
 t . \ dz 1 
 
 .Let z ss -i — - = - 
 
 v at; it 
 
 
 
 dw ^ a*w d# 
 
 1 
 
 IT 
 
 dv dz dv 
 
 ** </Sv 
 
 -l-i, 2 
 
 - 1 
 
 
 
 v:-a- 
 
 •)■' 
 
 
 3 
 
 . 5.-? 
 
 .-. U r= — sin- 1 — £: 1 
 
 2 
 
 = — sin" 
 
 .,3— 2v 
 n/5 
 
 3* - 2 
 
 = — sin-' — — : 
 
 V5* 
 
 = — sin" 
 
 ., 3x + 1 
 
 ( 3 x + 1 
 = tan > ( , 
 
 \2V1 + #- 
 
 ,>■ 
 
 
44 EXAMPLES ON THE 
 
 •™ n du 1 
 
 (20.) -y- = , 
 
 «* (# + b) \J x + a 
 Let # + a — sr ~r = *#> 
 
 dftt <#M Jo? 
 
 1 
 
 <#£ <&0 d^f (# 2 — 
 
 -a + &) ^ 
 
 2 
 
 . it = tan-i 
 
 2 
 
 2* 
 
 z> + b - «*' 
 
 */b — a >Jb — a 
 
 2 
 
 V& 
 
 # V 6 — a 
 
 ( 9i.) *U ! r 
 
 ^ (# + b) (x + a)« 
 
 = _J_{_i I-} * 
 
 a _ b ix + b x + a) ^ x _j_ a 
 
 a— b (x + b)»/x + a « — # (x + a)l 
 
 2 _! /x + a 2 1 
 
 = ^ tan A/ r 2_ + -7= 
 
 CJ — ^V v 6 — a a — b vx + a 
 
 ,_ v c?w 1 /-^ s x 
 
 o o 
 
 dx x ^ ' x y/^2 _^2 
 
 a 2 
 
 V# 2 — a 2 # >/^ 2 — a 2 
 
 •. u = ^# 2 — a 2 — a sec "• 
 
INTEGRAL CALCULI 45 
 
 <*>=-* 
 
 To integrate 
 
 **% %~ x'W d l Hfc x 2 V 'a 2 ± 2? 
 
 1 
 
 # 2 Va 2 ± x l 
 
 T * ^ — * 
 
 rfu dx da 1 #" 
 
 ^ dz ' dx ~ ^ *J drz 1 ± 1 ~ Va'V ± l' 
 
 J j rfz = -H — + log (x + ^' 2 + # 2 ), 
 
 or, = sin- 1 -, accordingly as 
 
 X CI 
 
 we use a 2 + a; 2 or a 2 — a 2 . 
 
 cm.) £- 
 
 ^ \/a 2 + f-Va 2 -^ 
 
 ^a 2 + a? 2 + ^a 2 — a? 2 
 3#* 
 
 1 vVj-^ 2 1 Va 2 — x l 
 " 2 tf 2 2 aJ 2 
 
 By last example, 
 
 = — Zx 
 
 + g lo 8 (* + ^«" + ^') - o 6in '-• 
 
46 EXAMPLES ON THE 
 
 (25.) — = A/ —I— = + 
 
 rt» / .. . 
 
 •\ n = a sin- 1 - — sj <# — # 2 « 
 
 (26 
 
 v du __ 1 As + a 1 
 
 dx x V a; — a "~ a/# 2 — a 2 
 
 ,\ ^ = log (a; + */# 2 — a 2 ) + sec- 1 — 
 
 * dx 'V a—x vV — «» vV-ar 5 
 
 To integrate 
 
 */a 2 — x 
 
 p = x 
 
 dq x 
 
 dx ~~ Va 2 — x 2 ' 
 
 | = 1 im.-4&ZW. 
 
 .-. f -^--^-' = - x*/ g> -- x 2 + f 'JcP-x'-dx 
 , = - s Va 2 -^ + o / / 
 
VL CALCULUS. 17 
 
 im p x*dx 
 
 p xdx p x dx 
 
 \ u = a I - + / — =; 
 
 sj2 op / X \ 
 
 (2R) - = * 
 
 ' ! * y/l — %acx + arc 2 sf\ - 2ac"-^ + a 2 c~ 2 
 
 Let 1 — 2ac# -f aV = 2 w — = 
 
 — Sac 1 # = ^ — = ~ 
 
 1 — %ac 1 x -{- a 2 c 2 ; 
 
 c 2 + a 2 4- 2 2 — 1 — a 2 c 2 
 
 n* 
 
 c* 
 clu du dx z c 
 
 dz dx dx ac z yV- - (1 — a 2 — c 2 + aV) 
 
 _^ 1 1 
 
 "" a ^/V — (l - a 2 — c 2 + a 2 ?)' 
 
 .'. w = log{* + \J z 2 — (1 — a 2 — c 2 + aV)} 
 
 = - log {>/l -2ac#4-a 2 c 2 + s/a 2 - 2acx + c 2 } 
 
 rr.r,*, dU 1 _ 1 
 
 (29.) T- = , Let a = - 
 
 J« (1 — « s )v/l +* 2 * 
 
 ill 1 sp z 
 
 ** = ^ (*» - if s/T+3 m ~ (*•— l) \/T+l? 
 
 Let 1 4- .; 2 = *» *■ - 1 = »* - 2, ^ = , * ■ 
 
 « v/v 2 -l 
 
48 EXAMPLES ON THE 
 
 du du dz _ \J v 
 
 1 
 
 dv dz dv (v 2 — 2) v +/v 2 — 1 v 2 — 2 
 
 du _ 1 C 1 1 *> 
 
 , M = 1 log Lt^*. -J- logil+^l: 
 1 . s/\ + * 2 + v 7 3 
 
 1 , a/ 1 + ^ + \/^ 
 
 log 
 
 
 d* (1 + #)s/#— 1 
 
 Let*»-1=* l+*W + », jj-^-f 
 
 du du dz \/v 2 + 1 
 
 •'■Tv~dz'da> («*+»)«• y/e' + l * 3 + 3 
 
 1 , • 1 t _j\/^ 3 - 1 
 : tan -1 —7= = 7= tan ~ 
 
 v/a v/8 \/s \/a 
 
 1 * -1 \A ~ »* 
 
 = 7= tan J ■ ■ • 
 
 v/2 V 8 * 
 
40 
 
 (3 1 .) / . — . ! X ■=. , dx = ; 
 
 J»S/X* + X — I * 
 
 /* dz p 
 
 4 
 
 1 
 
 -(•-i) 
 
 2 . ,9*— 1 
 
 . . 2 — a? . . m — 2 
 
 — sin" 1 — — — = sin * 
 
 
 (3o.) /- , Let \ + « - i 
 
 2 1 
 
 1 + x 1 = 2 - - + -, 
 z 
 
 dz 
 
 /» dx s* z 1 
 
 p dz __ 1 /* dsr 
 
 " J y/%Z i -%Z + l" "V^ // _1\ 2 l" 
 
 1 ■ f Vl +«* 1 — a; \ 
 
 = V^ ° S lv/^(l +.r) + 2(l +^)J 
 
50 EXAMPLES ON THE 
 
 7^ l0g { WT*) { 
 
 JLl ^(1 + *) 
 
 "va og i-^ + >/2^/r+? 
 
 . J_ log 8 (1 + *) {1 - * - ^3 s/lTtf} 
 ^2 g 1 — 2a? + a? 8 — 2 — 2ar 
 
 1. (2(1 -a?- x/2 x/l + a? 2 )l 
 (33.) /I ^ a . . Let 1 ^ rf = »V 
 
 »• = •• > , , 1 + ar = 1 . o 
 
 21og# =-log(s°- + 1)| 
 
 "*# == ~~ S 2 + 1 
 
 : ~ V2 C0 V2~~ v^ co 3* 1 
 
 + * a 
 
 Let = cot* 
 
 -^-av/L 
 
 */2# 
 
 cot 6 = 
 
 cos y = 
 
 cote _ VSa? 
 
INTEGRAL CALCULUS. 51 
 
 cos 8 = V rr5' •'• 6 - cos_1 Vrf5 
 
 y" </# _ 1 _, /l — ar 
 
 (34 - } jfc 
 
 <f^' 
 
 (i-^)v/(i + *•) 
 
 Let 1 + %"■ = *-« 2 
 
 1 0? 
 
 *» - 1 «- - 1 
 
 2 log # = — log (z 1 — • 1 ) 
 Jo? J# 
 
 ** — 1 
 
 dz 
 
 'J (l _ x l )xz ~ J z 1 — 1 *" — 2 " i/s? 
 1 . (v/iT^+n/2 ^) 2 ^ 1 y/T^+N/to 
 
 2 vs og i + ** - a* 1 <v/a ° s v/rr^ 
 
 C35 ^ / / ' t ±r. Let 1 + x = - 
 
 on = 1 </# = r 
 
 I 1 ° 11 
 
 L_0_4fcsi — +t — i + - — 1 = 1 + r 
 
 # z z z 
 
EXAMPLES ON THE 
 
 y^ dx p dz 
 
 Z X z 
 
 r dz _ r rU 
 
 n 
 
 = ~ log fs/^ + «_!+« + - 
 
 = ->«{^ 
 
 y/l — # — # 2 1 , 1 \ 
 
 + # + 1 + a 2 J 
 
 M ^r+io / 
 
 f 8(1 + *) 1 
 
 (- 2(1 +^)(^ + 3)-2\/l-^-^) l 
 D I a 3 + 6x + 9 — 4 + 4* + 4« 8 y 
 
 , 2 (1 + x) (x + 3 — 2 n/1 — * — or) 
 
 = l0 " — - — 6irr*F 
 
 — — / 7 — _L_r >. = COS"" 1 — . 
 
: ORAL CALCULUS. 
 dx 1 d z 
 
 IF 
 
 /* d x 
 
 , ~ 3 . Lets=-, dx = - 
 (a + bar)z z 
 
 /dz /-» z dz 
 
 « (as 2 +4)i «(« + J« 2 )4 
 
 /xdx ^___ f* 
 
 * ,— — = - as/ T-ar + y */T^a?d* 
 
 /» eta 
 
 = — Wl— a?* + / 
 
 */ V 1 — 4T 
 
 1 2 
 
 » /i s 1 
 
 + s sin - 1 #, 
 
 sin -1 x 
 
 = |«n- r « — (»-+») ^ J 
 
 
 9 
 
54 EXAMPLES ON THE 
 
 (39 .) f^dSL m*** 
 
 2 2 -4- i 
 
 # = # 5 , dx — -z dz, N /i» = ^ t » 
 
 /Jxdx __% r> dz __% • ~i z 
 
 a* 
 
 3 . .a* 
 
 3 
 
 X$ . A X 1 
 
 Let = sin"" 1 ~, sin = — , 
 a 5 a 5 
 
 A S1I1( 
 
 tan 6 = 
 
 Vl — sin 2 Va 3 — # 3 
 
 (40.) /* ^ j = fl dx 
 
 J {%ax+aFf J{(x + af -a 2 ) 1 
 
 Let 
 
 x + a = 
 
 -, d# = 
 
 — 
 
 dz 
 
 ~z* 
 
 
 
 = 
 
 -/-5 
 
 *» 
 
 ' 1 
 
 3 — a 
 
 f 
 
 = • 
 
 "At 
 
 ~« 2 ^ 2 )* 
 
INTEGRAL CALCULUS. 55 
 
 J_ 1 _J_ ___} fl 
 
 7 \/ 1 — a ' «" ™ a" . / . a* 
 
 + fl 
 
 
 ^ P% zdz 1 /* </# 
 
 ... /I— if = ~^WaI^ -^^wTi, 
 
 = — ( — jv/^aa? — » 2 . 
 
 //i i -•■* , px^dx rx* dx 
 
 n a? dx /» . ac 3 <?# 
 
 = — T7 n/ 1 — ff ' + CxsjY^x k dx 
 x 1 /• x <fs •* x 5 rfa; 
 
 ~ H + J >/TZT? "" y TT^T 4 ' 
 
50 EXAMPLES ON THE 
 
 ' J ,j i _ ^ - 4 iJs/l 
 
 # 2 1 
 
 — — " "j" \/ 1 — x* 4- - cos ""* (— # 2 ), 
 
 LetO = 7 cos- 1 (-a 2 ) 
 4 x 
 
 cos 4 = — # 2 
 
 tan 2 = A A- 00849 = A /I±? 
 "V l+cos4 A/ 1-aP 
 
 a 2 y> - « s y* = « 2 - c 2 x % 
 
I 3BAI I ALCULUS. 57 
 
 Xdx ~ (c'-yj 
 
 (I x a 1 ?/(l — c 2 ) dy c 1 — if y (1 — C*) </ ;/ 
 
 •" = -y? * « j (i-2/o~ {i-?){*-fy 
 
 /dx /a- — c- a; 2 /*i/ 2 (1 — c J ) rfy 
 
 7V a* - ^ ~ J (1 - f) (c s - f) 
 
 1/^1 1/^1 c r \ c r 1 
 
 r " Sy i +y " V i -y + %J e +y + a 7 c-y 
 
 1 i 1 — y , c i c + y 
 
 (44.) /it] '* 
 
 /I + x ~ 2 dx 
 
 Z — X~ l sjtf _ a + jy-2 
 
 /» d (# — a? "" 1 ) 
 
 >/(«-> — «)* — (a - 2) 
 
 1 , a" 1 — # 1 , a^ — 1 
 
 sec" 1 —= = — sec 
 
 \/a— 2 s/a — % y/a — 2 afJa — % 
 
 1 cos- 1 *^"-3 
 
 >/a - 2 # 2 -l 
 
58 EXAMPLES ON THE 
 
 FOKMUL^E OF REDUCTION. 
 
 y^ x n dx 
 \f 2 a x — x l 
 
 y-> x n dx rx n ~ l (a — x)d x n x 1l ~ l dx 
 
 y / 2 a x — x' 2 J s/% ax — x 2 J \J % ax — x l 
 
 ,, , , ', (a — x) dx 
 
 Makings = x n ~\ dq = \ -. r 
 
 y/S ax — x 2 
 
 and .*. ^ = (w — 1) # n ~ 2 <##, # = \/2 ax —x 2 , 
 
 we have fpdq = pq— fq dp) .\ f—p^====== 
 
 J \J % ax — x l 
 
 = a?- l <y% ax — a 2 — (ft — l)fx n ~ 2 \/z ax — x 2 dx 
 
 ==a; M - 1 V2 aa — ar — (ft— 1) . 2 a / _^ =r 
 
 */ \/2 ax — w 1 
 
 x n dx 
 
 mS r x u dx 
 
 Substituting this value in the original expression, we have 
 x n dx 
 
 y° x dx 
 
 /£ = T3 === ~ ® n ~ l \/% ax — x 1 
 \J % a x— x~ v 
 
 / .\ /* a^" l dx r x n dx 
 
 + Sa . (ft- 1) /-—==-. (n-I) /L_ == 
 ^\/3a«--^ J \J % ax — x 
 
 r x 1l ~ l dx 
 + a / 
 
4L CALCULUS. 
 
 (\ /• #"//./■ . / , 
 
 1 -f n — 1 / — ■ = — a;""" 1 V 2 ^-r 
 
 of 1 " 1 da 
 
 + a(3n- 1) / 
 
 y-* x a x 
 v/2 a # — 5" 
 
 a(2n-l) ^_j^_dx_ 
 
 /ar~ l a x 
 V^Tax — 
 
 n «/ v x a x — x 
 
 y^ x n dx r* n _ l (a -}• x) d x n x n ~ l d x 
 
 n/2 ax + x l J >/2 a #+#- */ \/% ax + a? 
 
 This becomes, if p = a?"" 1 
 
 /a?"- 1 (a + #) 
 
 (a + a?) rf# y 
 
 \/ ■* ax ~j~ x 
 
 rx"~ l (a + x) 
 
 _ ( n — }) f aT'ty/si ax +x 2 dm sb ar-*y/<i ax + .r 2 
 _ 2 . a (n - 1 ) / ■ - (n - 1) / , 
 
60 EXAMPLES ON THE 
 
 Substituting this in the original expression, 
 
 x n dx , / ^ '* , n x n ~ l dx 
 
 r> x ax , / •* N /» i 
 
 / , r = #"- 1 v / 3aa + ^ — 2a (n— 1) /— 7 
 
 J sj)lax + x 2 J sj 
 
 %ax -\- x 2 
 x n dx /» x n ~ l dx 
 
 A /* # >l a# /» x n ~ l dx 
 
 (n - 1) / , : - a / — 
 
 *J \J %ax + # v J \J fyax+a? 
 
 f~ l sj^ax + x 2 a(2w — l) /■• x n ~ l dx 
 
 J s/^ax + x 2 
 
 J x 
 
 n 
 dx 
 
 x n sj% 
 
 ax — ar 
 
 d# __ p dx(a — x) 
 
 ' x n ~ l J%ax — x 2 J x 
 
 x 1l ~ l \J %ax — x l J x 11 \J%ax — x 2 
 
 p ax 
 + a / 7 f a .) 
 
 J x n J%ax-x 2 K } 
 
 ^ T , . 1 , , (a — x) dx 
 Now, making p = — - and a^ = ■ . , 
 
 x \J%ax — x 2 
 
 and .*. dp = — ttaT*~ ! d#, q = \/%ax — a? 2 , 
 
 there results 
 
 /• (a — x )dx __\/ % 2ax — x l n\J Sax — x*,dx 
 
 ' x n <s/%ax-x 2 "" 2? + V ^ 
 
 \/%ax — x 2 r _^ 
 
 = — H + San I 7=== 
 
 x n J x n \J %ax — x' 
 
 r dx 
 
 — n / - 
 
 J x n ~ x sJSax — x l 
 
61 
 
 Substituting this in the equation (a) we obtain 
 
 y~ dx sJSlax — x 1 p dx 
 r — / = — 2 an I . 
 &- y/Xax — x 1 * Jx n \/2ax— a? 
 
 P dx p dx 
 
 + n I + a I , 
 
 J x n ~W%ax — of J x n s/%ax - x l 
 
 /n «\ r dx y/Zax—x* 
 
 .*. a (2m — 1) / = 
 
 J x n *J%ax-x- n n 
 
 + (n-l)/* *JL „ 
 
 Jx n -W2ax-x' i 
 
 y'_ j*fj %/% ax — x l 
 x n <s/2ax — x l a(2w — l)o5" 
 n -— 1 /» <£# 
 
 /^ Jo? 
 
 y" ^ __ f (a + x) dx n dx 
 
 x n ~ x \/%ax + x 2 J x n s/2ax + ar 2 */ # n V2a# 
 
 + ** 
 
 ™ -» 17 (« 4" #) d# 
 
 11 jp = # ■ and dq = v — 
 
 s/ % 2ax + a?' 3 
 
 i r(a + x)dx _ s/%ax + x 2 Ps/Zax+x^dx 
 
 J x n »/2ax + x~ x n +w/ a"* 1 
 
 _ •SSax+x* P dx 
 
 x" ' an J x n *J%ax + x 1 ' 
 
 p dx 
 
 + n I— . =, 
 
 J x n ~'s/%ax+ x l 
 
 and by substitution in the first expression there results: 
 /» dx __ */Xax-\-x : 
 
 * ■ .c"- 1 n/ 2 a a; + x l x" 
 
62 EXAMPLES ON THE 
 
 . /* dx p dx 
 
 + (Zan — ft) / — + n / =, 
 
 (1 — ?i) /L__^ */%ax + x 2 
 
 a(Zn — 1) JaT-Wjlaat+x* x n a(2n-l) 
 
 p dx 
 
 J x n *S%ax + a? 2 ' 
 
 /T-s ^r« We have , ., — — 
 
 1 a 2 1 f <r + a? 2 _Jl_! 
 
 "" €? ' (ft 2 + «')" "" <? l(« 8 + «T ~~ (^ + »')"/ 
 
 1 a; 2 
 
 a 2 (a a + «*)"- 1 a 2 (a 2 + #*)"' 
 
 1 px*dx 
 
 /dx ^ P ^ x 1 PA 
 
 ifp a ^(lp ( ^ ; 
 a: 2 c?# 
 
 2 i «^»-l 
 
 /* a:'ftaj — x 
 
 "Jiff + **)* = (2n — 8) (ft 2 + # 2 ) 
 
 1 r» <## 
 
 + %n — %J (ft 2 + a?*)" 1 ; 
 
V (a 
 
 INTEGRAL CALCULUS. 63 
 
 d x 1 x 
 
 2 + x z ) n a 2 (»n - 2) (a 2 + x l ) n ~ l 
 
 dx 
 
 l r dx i /* 
 
 Also for / 7-^- — sr= = i x m ~ l dx - i . We have, 
 
 J (a 2 + or)" % / (a- + #«)" 
 
 if f = ^" , - 1 , rf? = # (a 2 + <*)-" ^ : 
 
 .*. XL 
 
 n x a x 
 J (a 2 -f xy 
 
 1 x m ~ l m — 1 p x m ~ 2 dx 
 
 "" 2— 2» (a? 4- ^" 2 ) n - 1 + w* — 2 */ (a 2 + a? -)"" 1 
 
 Ju = (a 2 — # 2 ) a dx 
 (a> _ ^y = a 2 (a 2 - a; 2 )" 1 " - & {a 2 - 1 ) % \ 
 = o? / (a 2 — x*) 2 dx — / x . x (a 2 —< x 2 ) * <f 4 
 
 N 
 
 . n •zJ x(a 2 — x 2 f u 
 
 = « 2 (a 2 — x 2 ) ■ da; + -i £ ; 
 
 J v n !• 
 
 »» 
 •. KB U —I 4- - (a 2 —x 2 ) ■ eta 
 
64 EXAMPLES ON THE 
 
 r x n dx ^ a, "V2flg-^ 
 J \f i &ax—x l ~' n 
 
 /%n— 1\ /» tf n-1 dx 
 a \ n Jjy/Zax — ap' 
 
 If we make n = 2 this expression becomes 
 
 Also / — ^ = — v/^aa; — a;' 4 + a ver sin -1 ( - ) ; 
 
 J s/Kax-x* ^ aJ 
 
 x 2 dx x x/Sax — x* a /- 1 
 
 /# — . = z. — _- ^ %ax — x l 
 
 s/^ax — x" * % 
 
 Sa 2 . ,/x\ 
 
 -j- -r- ver sin ( - ) 
 
 2 \aj 
 
 (x + 3a) /- x , 3a* . _ x fx\ 
 
 = — i — s/2ax — x % + — - ver sin 2 f - 1 
 
 n dx I ?L__. a 
 
 y (a 3 + <r = 2w — 2 # a 2 (a 3 + a?*)— 1 + 
 
 Sn — 3 1 r__^___ 
 2^T~2 ' «? 'J{a 2 + x v ) n - 1 ' 
 
 If w = 4 we have /-rs-: — ^r 4 = 
 ^7 (a- + # 3 ) 4 
 
 J_ 
 
 6 a 2 * (a 
 
 1x6 r dx 
 
 67 3 ' (a 8 + a; 2 ) 3 + 6a\/(a 2 + >)*' 
 
1NTJ I.CULUS. 
 
 y» dx 1 x 3 /• eta 
 
 A1 /* ' /r X ^ . 1* -1* 
 
 .*. the required integral is 
 
 x o 
 
 6 a 2 • (a 2 + ^ 6 . 4 a 4 (a 2 + ar) 2 
 
 5 . 3 a 5 . 3 J_ -i/^ 
 
 + G . 4 . 2a 6 ' a 3 + # 2 + 6 . 4 . 2 a 7 ' an \a)' 
 
 ydx 
 
 The formula of reduction is 
 
 1 (ar + 1 )* n — 2 /■ rf * 
 
 (n - 1) &~* ~ n— lj x n-2 ^ + ^)f 
 
 If n = G, we have / 
 
 J x r > (1 + <*)J 
 
 1 (a 2 + 1)* 4 f__Jf_ 
 
 /■ <* a _ 1 (a^ + l)* __ 8 /• </ .r 
 
 */ *< (1 + tf)i " 3 ' *> I J f ^ + , ).</ 
 
 /• </#_ — (1 + a 2 )* 
 
b EXAMPLES ON THE 
 
 And by reduction this becomes 
 
 r dx _ 1 (l+'jfr)* 4 (1 + x*)\ 
 
 J sfi (»* + I)* "" & rf + 3 . 5 ** 
 
 4. 2 ( 1 + x 2 )* 
 5.3'"" x 
 
 dx 
 
 J x 1 
 
 vV-i) 
 
 The formula of reduction is 
 
 ydx 1 (a? 2 — 1)4 ?i — 2 r_dx_ 
 
 x n (x 1 — 1)4 ~~" w — 1 ' # w-1 » — lJx n ~ 2 (x 2 — 1)4' 
 If n = 6, this becomes 
 
 /» </# _ i (# 2 — i)4 £ /2_^1___ 
 
 c/ ^(^-l)T"5 ' ~"""^ * 5\/ **(**-- 1)4 ' 
 
 r dx 1 (a? 2 — 1)4 2 r dx 
 
 also / == • * - J_ — I 
 
 %/ x*(x 2 - 1)4 3 a 3 T 3 t / ^(tf _ x)4 
 
 r dx _ (ar — 1)4 
 
 */ a? 2 (V — 1\4 
 
 a* (a? —i)i « 
 
 Therefore, we have the required integral 
 
 - 1 fo 2 - 1)* 4 (a 2 - 1)4 2.4 (a* --1)4 . 
 ""5* a; 5 + 3.5 ^ + 3.5 x 
 
INTEGRAL CALCULUS. 
 
 67 
 
 CHAPTER IV. 
 
 (1.) ^L=x*(lo g xf (log*)'^, ^| = ^ 
 
 
 # 3 (log a:) 2 2 
 
 U SB ^° 
 
 — - /x 2 dxhgx 
 
 I x 2 dx log a? 
 
 i da \ 
 
 dp I 
 
 dx x 
 
 9f 
 
 x* log X 1 
 
 -if' 
 
 dx 
 
 __ a? log a? 
 3 3 J ~ ' = 3 "9' 
 
 # 3 (log#) 2 2# 3 log x 2# 3 
 .'.* = 3 — + — 
 
 = ^{(log^) 2 -5log^ + ^}. 
 ^ x^dx 
 
 = j(log *)-* + jyij /**• (log •)"« 
 
 7> = (log »)~" i rf<? = a -1 i * 
 
 j 1„ s_i( <*# «' 
 
 q =v 
 
68 EXAMPLES ON THE 
 
 Similarly integrating by parts, 
 
 y~* 3 * 3 3 /"* 5 
 
 afdxQ.ogx)''* = -r-( lo S ^)~ 5 + Z — 7 /*W*(log*)~^ 
 
 hfidm (log*)~* = j(log*)~* + _. p^dx{\o%x)~\ 
 
 *\Z*(log a?) "« = j (log a?)""* + __ tfdx (log*)— 2, 
 
 4 \/log * 
 
 t* = 
 
 4v/lo^ t + 8 log* + (Slog*) 2 + (Slog*) 3 + C, J # 
 
 dx 
 p x x dx /» 5 x 
 
 ®->J Q^Hf-J X (log*) 8 ' 
 
 p z=. x* dq = 
 
 <#^ = hx*dx q = — ■ 
 
 dx 
 
 x 
 
 (log *) :i 
 
 1 
 
 2 (log*) 2 
 * 5 5 /* **d* 
 
 = ~" 2 (log*) 2 + ^y (log *) 2 ' 
 
TM ! no 
 
 In like manner, 
 
 „ prfdx 
 
 y* ar x* p 
 
 2 (log xf = "Io^ + V 
 
 2 (log xf lug x J log x y 
 
 & $& 25 rx x dx 
 
 2 (log xf 2 log x T % J log 
 
 25 rx x dx 
 x 2 ,/ log #' 
 
 £ = rf ^ = a* J 
 
 I c?p =3 # 2 J# q = — 
 
 a; 3 a- 1 3 /» . 
 A Aj 
 
 / a* or rfi» b= I a x x ax 
 
 r t a's 1 /*' , a* x a* 
 
 J A ALA A \ A A /J 
 
 r# 3 s* 8 ?>x o ■> 
 
 (5.) /or* (Jog xfdx p — (log a?) 3 dq = x*dx 
 
 x 4 
 
 /"* &'* 3 ' / * 
 
 / x* (log xf dm = — (log xf — - i x* (log »)■ efa 
 
 # 3 (log a) 1 tifa — — (log.r)" 2 — - I x ] log, 
 
70 EXAMPLES ON THE 
 
 / a? dx log x = — log x — - / a? dx 
 x* . 1 4 
 
 .\ I # 3 (log#) 3 dx S3 
 
 ^ 4 f ,1 ^ 3 /i no , 3.2. 3.2\ 
 
 — | (log #) 3 - - (log #) 2 + — log x - -r- J- 
 
 (6.) / e x x*dx = e r x* — 4 A r ar 1 da: 
 
 /V a? 3 <f * as e* a; 3 — 3 A' a? 2 cfa 
 / ^ # 2 £?# = e* # 2 — 2 / e 5 a? dx 
 I e x x dx = e* a — / e x dx == e x x — e x , 
 .-. Ce x x x dx = e x (x* — 4# 3 + 12# 2 — 24a; + 24). 
 (7.) Le- X x % dx = — <r-*# 3 + 3 /V*# 2 Jo; 
 / e~ x x 2 dx = — £-*a 2 + 2 / <?-*# dx 
 je" x x dx=z — e- x x + / e~ x dx = — e - ^ — $—*, 
 . . re~*x i dx=— e- x (x i + 3x 2 + 6x + 6). 
 
INTEGRAL CALCULUS. 71 
 
 (—1 . *£ * \ 
 
 = v/IT^iog i f-/" v/l ^ r ^^ 
 
 - yrr^iog, - A^= - Aj£= 
 
 J x s/l+ap J %/\ + x l 
 = v/'l + x 1 log# — log — — — x/l + J? 
 
 = v/l + 5 (logo? - 1) - log 
 
 1 + V l + *" 
 
 (9.) / f rf* ?_ = / ^ <te -7 -T\— 
 
 =/-{ ( ^-"(S4)} 
 
 = «* =■• 
 
 0+1 
 
72 EXAMPLES ON THE 
 
 (1 0.) /** — i / = / e v dx 
 
 J (1— ar) Vl — a* J (l-ir)Vl r ^ 
 
 a/ I n/ 1 _ ^ (1 — a?) n/1 — a^J 
 
 J [vrr# wi — «/j 
 
 xe* dx ., e r d x 
 
 dp = dx q = — — —, 
 
 J 2(e*— l) 2 
 
 /#£* dx x 1 r dx 
 
 («*— I) 3 2(V-1) 2 + %J{firmXf 
 
 x 1 r dx 1 /» e* c?# 
 
 = 7J 2(^-i) 2 7" 2^/^ - 1 + %J(f — \y 
 
 x 1 Pe x dx 1 re* dx 1 /* e'efo 
 
 """"2(^-1)2 + 2c/ "e* 2^/6* - 1 + 3i/(«* - 1> 2 
 
 a? 1 g* 1 1 
 
 ~ ~ 2(e*- I) 3 + 2 ° g s* — 1 ~ 5 *• — I 
 
 - § °8 ^r _ l 9 («■ — 1) t + (? - 1 J 
 
INTEGRA] < m.< i-j 
 
 (In 
 
 (12.) — = — /> = a* JL=: 
 
 dx dx x k 
 
 dp 
 
 a* t A pa x dx 
 
 — = Aa* q = — ■— ., 
 
 dx 1 3#* 
 
 a* a pa* di 
 
 ■-■ u = -s*> + Hj — 
 
 Similarly, 
 
 /a T dx___ a x A Pa'dx 
 ~x r "' ~ 2^? + "2*7 ~^~ 
 
 (13.) — = — ^ = * — =SO», 
 
 dp _ 1 3 a* 
 
 .-. u = 1 H / # *a T dx, 
 
 A* 1 2A«^ 
 
 /' — 3 « r 3 /* — ^ 
 ) / m *a*dxwa — - h / * -a*</# 
 
 ^ A or* %aJ 
 
 /x~*u x dx= — ;■ H C x~~ ll a T dx, 
 
74 EXAMPLES ON THE 
 
 1 ( a 
 
 3 / a* 5 a* \] 
 
 A#2 ^ A A# 
 
 o» ( -, , 1 3 3 . 5 p 
 
 Another result may be obtained by the following method. 
 
 = x ^ 
 
 dq 
 
 d® s/x P a dx 
 
 ' x 
 
 d p 
 
 = Aa x a = 2 xh> 
 
 d x 
 
 Also, ^/ a x x 2 dx= -a x x* — - A / a x x*dx. 
 
 / a* #1 d a?sr - a* s« — t A / a* x* d x, 
 Ca x x* dx= - & s£ — s A^/ a* ^<J#, 
 
 sb a a* «' — a AS - a* x* — - A ( - a x a? 
 [3 3 \5 
 
 2.2 7 > \ 
 
 — - A - a* #*, &c. ) 
 
 .*. u 
 
 a x r 
 
 Att=s ~ J 
 
 A V # t 
 
 (&r A) 2 (2 a A) 3 (8 a? A)* 
 
 2 # A — v _ y + 
 
 ' 3.5 3.5.7 
 
 + &c .}• 
 
INTEGRAL CALCULUS. 
 
 d u 
 
 (11.) '— = x> 1 *& = x m \ 1 + n x log x 
 
 n 2 
 = ftf" + »«•+> log x + - *•+' flog #) 2 + ftc., 
 
 + [^ y 1 * m+2 ( io s^) 2 ** + *c. 
 
 x m (\ogx) n dx= ) (loga?)* 1 (logs)*- 1 
 
 n(n-l)(n-g)...3.1 j 
 
 # ,n+1 log#d#= - j log # 1 
 
 *"+»(log*)'rf*= f-_ {(log*)* 
 
 x m +* Qogxfdx= \ (log xf - (logz) 2 
 
 1 G ^ 
 
 Arranging the terms according to the powers of log w % 
 
 B 9 
 
76 EXAMPLES ON THE 
 
 r ,7_* m+l nar+ ' 2 n2 * m+3 n3 * w * 4 » * 
 
 f ^ +2 wa m+3 w 2 tf*+ 4 , ? 
 
 -f n log x < — + —, &c > 
 
 5 lm + % (ro + 8) (m + 4) :i 3 
 
 + — ^ •>" isr+8 - osm? + H 
 
 If ^ = all the terms vanish. 
 If x = 1, / x nx ,x m dx becomes 
 
 n 
 
 1 L See. 
 
 m + I (ro + 2) 2 ' (m + 3)* (m + 4) 4 
 , ' . tfw logs: . dq 1 
 
 (15.) -r— = — -^ » = log# —2=--; - 
 
 k J dx (l+xy * 5 dx (I + x) 2 
 
 dp 1 
 
 log # 
 1 -|-# 
 
 dx x \ + x 
 
 /dx 
 * (1 + *) 
 
 log # r d x r> dx 
 
 ""* r+~# y *# y i +* 
 
 log 0? 
 
 1 4-# 
 
 4- log a; - log (1 + #) 
 
 1 + r 
 
 log x — log (1 4* x). 
 
INTEGRAL CALCULUS. 77 
 
 CHAPTER V. 
 
 )dS 
 
 (1.) f sin* Qd8 = /"sin 8 (1 - cos 2 6) 
 
 = / sin 9 dO — / sin 9 cos 2 9 
 
 de 
 
 A cos 3 cos 9 (1 — sin 2 9) 
 = — cos 9 h — == — cos 9 H — - ; 
 
 o o 
 
 2 . cos 9 sin 2 9 
 = _-cos9 __ 
 
 (2.) fcos 3 QdO = /cos 6(1- sin 2 9) <J0 
 
 = /cos 9 d 6— /cos sin 2 S d6 
 
 . A sin 3 9 . sin 9 (1 — cos 2 9) 
 = sin 9 — = sin 9 ^— ' 
 
 o o 
 
 k 2 sin 9 sin 9 cos 2 
 
78 
 
 EXAMPLES ON THE 
 
 . ft . rdO / (sin 2 Q + cos 2 6)^0 
 
 sin 3 
 
 J sin <9 ^/ 
 
 / A cos6d6 1 I /»sin 
 
 cos . q ■• = — cos ■ . „ — - / — r 
 
 / » dB __ 1 /» dB 1 cos 
 sin 3 e~~%J skT0"~"2sin 2 
 
 = i log ( tan - ) — - 
 , . . p sin 3 6d6 p 
 
 <*•> j i5Fr =y 
 
 cos 2 0^0 
 sin 3 <9 
 
 sin 6d6 
 
 \ 1 cos 
 
 sin 2 
 
 sin0J0 /» sin 0^0 
 
 cos 4 
 1 1 
 
 cos 4 
 
 -/ 
 
 cos 2 
 
 3 cos 3 cos 
 
 = 5 Th\o - 1 + sin2 H 
 
 3 cos 3 C3 J 
 
 _ _L_ / sin* - |l. 
 3 cos 3 8 1 3 J 
 
 (5.) / a? 2 tan~ l A/ - dx. Let - = # 2 
 
 x° = a° ar 
 
 a; 8 d[« = 2fl 3 # 5 rf#, 
 
 .-. / # 2 tan -1 a/ - dx = %a? j z 5 tan"" 1 #^2 
 
 jj = tan"" 1 # 
 1 
 
 dp = 
 
 1 +S 8 
 
 J# 
 
 rfj = 2 a 3 # 5 d # 
 
 3 ' 
 
EGBAL CALCULUS. 79 
 
 <**• , <£■ r z' 
 
 = _ ( *o 4 . 1 )tan-.*-- T {-- 3 -+l} 
 
 = a \— ) tan V i - rfr its* - 3-« + x J 
 
 (6.) — = cos 6 6 p = cos ^ --^ = cos 
 
 -^ = — . 5 cos 4 6 sin 6 q = sin 0, 
 .-. r cos c 0^0 = sin0cos 5 + 5 y\» s 4 sin 2 d 
 
 = sin 6 cos 5 -f 5 ^cos *6 dx— 6 fco^Bdx, 
 
 .-. / cos 6 Odd = - h- / cos 4 0</0. 
 
 Similarly, 
 
 /I 3 /» 
 
 cos 4 0d0 = -sin0cos 3 0+- / cos 2 W0 
 
80 EXAjMPLES ON THE 
 
 /l 1 /• 1 
 
 cos 2 0d = - sin0 cos + - I dS— -sin0cos0 H — 
 
 /„. sin0 cos 5 5 fl . . _ 3 
 
 cos 6 d0 = + - \- sm (9 cos 3 + - 
 6 6 (4 4 
 
 (- sin 0cos + a J? 
 . „ Ccos 5 5 cos 3 5 cos 0^ 5 
 
 = sm n-r + -2T- + -T6-} + i6- 
 
 (7.) — = sin 2 cos 4 = cos 4 0— cos 6 0. 
 
 But from preceding example, 
 
 30 
 
 8 
 
 /4aja • /, fcos 3 , 3 cos0) 
 cos 4 BdS =sm <— 1 — > + 
 
 /« /i ^/i • a f cos5 ^ . s cos3 # 5 cos 0) 5 
 
 . „ C cos 5 cos 3 cos 1 
 
 .*. % = sin < • . 4- H > a 
 
 X 6 ^ 24 ^ 16 J ^ 16 
 
 . - f cos 3 sin 2 cos 3 cos 3 cos 7 
 
 . . f sin 2 cos 3 1 . ,„ . „ ^ cos 0") 
 = sln ,| _ g cos*(l- S m^) +lr } + jj 
 
 sin 3 cos 3 sin 3 cos sin cos 
 
' 
 
 du 
 
 (8.) -j- = sin G cos 3 6 = sin G cos — sin 8 cos 0, 
 
 sin 7 sin . . 
 
 lt» — -- = sin 7 
 
 8-*9 
 
 . n i _ cos 2 01 
 
 { 9 ^ 63) 
 ( '' y sin* 1/ sin 5 "V sn? 
 
 cos 2 0dtf 
 
 /cos* Hi 
 sin 5 
 
 2? = cos 
 
 dp 
 \dl 
 
 dq COS ( 
 
 db sin 5 9 
 
 • sin # = — 
 
 1 
 
 4sin 4 
 
 /cos 2 0d0 _ cos0 1 r dB 
 
 sin 5 ~~ 4 sin 4 *"~ 4*/ sin* 0' 
 
 - r : 
 
 ♦3$ J$ 
 
 cos0 3 
 4sin 4 4^ sin 3 0' 
 
 cos0 1 p dB 
 
 /dB cos0 1 r dB 
 
 sin 1 ~~ ~~ 2 sin 2 + Hj smfi' 
 
 /dB cosfl I / 0\ 
 
 sln^ = -Slin^ + 2 l0g l tan 2>)' 
 
 H 
 
 = — cos 
 
 cos 3 cos 3 , / \ 
 f 1 8 I .1 , / 6 \ 
 
 ■ s 
 
82 EXAMPLES ON THE 
 
 r dB /- sm 2 e dQ r dd 
 
 ^ ' J t cos 6 6 ~~J cos 6 4 J COS 4 
 
 J^ sin 9 
 
 ^9 = sin i 
 
 <?0 COS 
 
 ~T7 = COS P 
 
 rf 9 * 5 cos 5 9 
 
 ''sin 2 ^6? (9 sintf I /^tf 
 
 cos 6 ~~ 5cos 5 5c/ c 
 
 cos 4 
 
 4 /» Jfl 
 
 M ~~5cos 5 5,7 cos 4 0* 
 
 Similarly, 
 
 / <#<9 sin0 % / d6 
 
 cos 4 (9 r= 3 cos 3 (9 + 3^ 
 
 cos 2 
 
 sin 2 sin 
 
 + r> 
 
 3 cos 3 3 cos 9 
 
 sin 4 f sin 6 2 sin 
 
 + 
 
 f sin 6 2 sin 6 1 
 5 cos 5 <9 ' 5 1 3 cos 3 + 3cos0j 
 
 _ . f 1 4 8 ^ 
 
 * Sm 15 cos 5 + 15 cos 3 + 15 cos 0J 
 
 (\\\ du . — sin5 ^ _ sin^ (1 — 2 cos 2 6 + cos 4 0) 
 ^ ' dd ~~c^"~~ cos 2 ^ 
 
 sin 
 
 cos 2 (9 
 
 2 sin + sin S cos 2 0, 
 
 I _ A cos 3 (9 
 
 ?* = H + 2 COS ^ — 
 
 cos0 3 
 
INTEGRAL CALCULUS. W 
 
 If, _ „- COS 4 01 
 
 = — A 1 + 2cos 2 <9 — [• 
 
 cosflj 3 J 
 
 If, ^ ^ . o* l-2sin 2 9 + sin 4 9) 
 
 = \ 1 + 2 — 2 sm 2 — — I 
 
 cos0( 3 J 
 
 1 f sin 4 9 4 sin 2 9 81 
 ~~ ~" cosl ( 3 + 3 ""3}* 
 
 ma\ du cos 4 # dq cos 9 
 
 (12.) -rxss-T-rs ^ = cos 3 9 -f= 
 
 v ; dO sm J <9 ^ r/9 sin 3 9 
 
 -^■= — 3 sin0cos 2 a = — ^ . ,, . 
 d0 * 2 sm 2 6 
 
 cos 3 3 /» cos 2 B dS 
 ' ~~ 2siir<9~~ 2 J sin0 
 
 cos 3 3 r cos 2 Odd 3 rcos 4 6 d8 
 
 /cos~ 6 dB 3 rcos A 6< 
 sin 3 (9 + 2^ sin 3 < 
 
 2 sin 2 2 J sin 3 (9 r 2^ sin :i 
 
 ___ cos 3 (9 r cos 2 BdB 
 
 U ~*i$? + 1/ sin 3 
 
 e?£ cos ^ 
 
 j} = cos 
 
 rf<9 sin 3 
 
 * ■ = — sin q = . 
 
 tf <9 J 2 sin 2 
 
 *cos? OdQ cos0 
 
 /cos 2 Bdv _ cos 6 \ r d6 
 sin' (9 "2siir^"" 2,/ siiu9 
 
 ±log(ten£} 
 
 COS0 
 
 2 bin 2 6 
 
84 EXAMPLES ON THE 
 
 cos 3 6 3 cos 6 3 
 
 sin 2 2sin 2 
 
 - 5 log(tan£) 
 
 1 f 3cos0) 3. / 6\ 
 
 4 i\M - 
 
 , v du __ 1 1 
 
 d S sin 2 cos 3 cos 3 sin 2 6 cos 
 
 sin 2 # 1 1 cos0 
 
 ' nna A "» AAQ /J "• oii.1 i 
 
 cos :i costf cos# sin 2 
 
 s* sin 2 6 dO ^ P d6 1 
 
 u = / «- + 2 / — - - — 
 
 ,/ cos° J cos 6 sin i 
 
 2? = sin 
 
 >sin 2 0d<9 . ^ r dB ^ 
 dq sin 
 
 d0 cos 3 
 
 dp . 1 
 
 "" ; = COS £ = j 
 
 /sin 2 6 dB ___ sin 1 n d6 
 cos 8 (9 '""" Scos 2 ^ ~" 2 */ costf' 
 
 ___ sind § r dB 1 
 
 "" 2 cos 2 ^ H J cos 6 sin 
 
 i ci — cos 2 ^ _-i 3, r /* e\\ 
 
 = s-i^iT^F - *] + 2 lo n tan U + §)} 
 
 1 f 1 3) 3, /« <9\ 
 
 = S*lw* - 5J + a log tan U + 5> 
 
INTEGRAL CALCUl 
 
 dO sin* cos 2 sin 2 cos- sin 4 <9 
 
 1 1 1 cos 2 fl 
 
 ' cos'- 6 + sin 2 sin*0 sin 4 0* 
 
 M = tan0 + 3/ T-T5 + / -^r 
 */ sm 2 */ sm 4 
 
 cos 2 0d6 
 T 
 
 j 
 
 dq COS # 
 
 ;; = COS -r^ = -: — r- 
 
 cos 2 0</0} i d0 sm 4 
 
 sin 4 ) rfo . . 1 
 
 1 3 sin 3 
 
 cos0 \ r dQ 
 
 
 3sin :i 3i/ sin 2 
 
 cos0 5 /» dO 
 
 San'l + ^ + s/sFa 
 
 cos 2 sin 5 cos 
 
 , + TTT*- 
 
 3 sin 3 cos cos 3 sin 
 1 — sin 2 3 sin 2 0—5 cos 5 
 
 3 sin 3 cos 3 sin cos 
 
 1 1 + 3sin 2 0- 5cos 2 
 
 + 
 
 3 sin 3 cos 3 sin cos 
 
 1 4 (sin 2 — cos 2 0) 
 
 1 1 8 cos 3 sin cos 
 
 1 8 cos 2 — sin 2 
 
 8 sin" cos 8 2 sin cos 
 
 1 
 
 cos 8 
 
 - cot 2 0. 
 
 85 
 
86 EXAMPLES ON THE 
 
 (15.) ^ = tan 4 9 = tan 2 (1 + tan 2 0) - tan 2 
 
 = tan 2 (1 + tan 2 0) — (I + tan 2 0) + 1, 
 tan 3 . A 
 
 .'. u 
 
 = — tan o -f- y. 
 
 o 
 
 
 -^ — rTv n 
 
 ('••) w ; 
 
 1 1 + tan 2 1 
 
 ""tan 5 tan 5 tan 3 
 
 
 1 + tan 2 1 + tan 2 i 1 
 
 
 tan 5 tan :3 ' tanO* 
 
 
 .1 J h 1 ~m r) 
 
 .*. 11 
 
 ~4tan*0 ' atan 2 +h ' Lsin9 - 
 
 <"■> 3? 
 
 = G 3 cos ?) = 3 -4 = cos 
 1 dQ 
 
 
 &-«? « = M 
 
 
 .-. u = 3 sin<9 — 3 r&&n6dQ 
 
 
 p = 6* ?|=sin0 
 
 fb 2 sm6dd = — <9 3 cos<9 + sfbcosO < 
 P = 6 g=cos* 
 
INTEGRAL CALCULUS. 87 
 
 /# cos 6 (IB = 6 sin 6 — /sin OdO = 6 sin 6 + cos 0, 
 .-. M a B's'md -f 30 2 cos0 — G ^ sin <9 — Gcostf. 
 
 ~ * du x 1 . 
 
 (18.) — = — sin" 1 x. 
 
 dx y/l—tf 
 
 m . x~dx da 
 
 To integrate — p — x — 
 
 sj\ — x l ' ilx \Zl—ar 
 
 dp 
 dx 
 
 -1 ^ = -^1-^ 
 
 /* — ,. ^ = — a? v/l — & + / y/i — x l dx 
 
 k ' sj 1 — X 2 J 
 
 x~dx 
 
 -_~,/i-r^+ r dx r *** 
 
 — — X \/ 1 — X -f-l — / , : 
 
 x 2 dx . _ 
 Ts ext, to integrate — ,. sin x a?, 
 
 dx sj\—x l 
 
 dp 
 
 x / r 1 . 
 
 (Z = — V 1 — *' + o sm w > 
 
 dx Jl—x 1 
 .-. u = - (sin" 1 »)* — - ^/ 1 — .^sin" 1 ^ -f - / *<** 
 1 /^eixi^adx 
 
 y/l —X* 
 
88 EXAMPLES ON THE 
 
 1,., ' X2 cos/ 1— x* a? 1 . 
 = - (sin l xy ^— sm 1 x + — (sm ! #)~ 
 
 1 > . _! vo a \A — # 3 • -i , ^ 2 
 
 = - (sm 1 xY *-_ sm 1 x + — - • 
 
 4 V & 4 
 
 (19.) — = -sm- 1 a?. 
 
 dx (l_tf 2 )§ 
 
 p = sm * # 
 
 dx (1—^)1 
 
 ^ _ 1 _ 1 
 
 sin"" 1 # /"• ^# 
 
 star 1 * 1 r /» dx t > r__^f\ 
 
 sin- 1 ^; . v/l — # 
 
 = , + log v , • 
 
 . <?m a? 2 , . 
 
 (20.) ^~=r- — i tan" 1 ^. 
 v ' d# 1 + x 2 
 
 i cfa # 3 _ 1 
 
 p = tan" 1 X ■=— &s - — ; r = 1 — ^— ; 5 
 
 ^ dx \ + x 2 1 + ar 
 
 d» 1 , _ a 
 
 — s= - q =z x — tan A #, 
 
 d# 1 -f- ar 
 
 #«?# /itaiT" 1 re <fo 
 
 . , . N p xdx pmx 
 
 .-. u = tan" 1 # (# — tan" 1 x)—i ■ * , -f / — - 
 
INT I CUB. 
 
 sss tan -1 a; (x — tan" 1 ^) — log \/l -f- ar + ^ \ 
 = x tan" 1 as — - (tan" 1 x) 1 — log >/l + x\ 
 
 ('21.) — = a« sin 2 a:. 
 
 (IX 
 
 dq 
 
 p = sm' x —±=e 
 
 ax 
 
 a i 
 
 o.ax 
 
 dp . . 6° 
 
 — = 2 sin a; cos ^ = 
 
 1 2 /* 
 
 n=-e ar sin 2 a' / tf a * sin#cos#dfa? 
 
 dq 
 
 p = sin # cos £ 
 
 
 '* 2 • 9 
 
 -— = cos 2 x — snr x 
 
 (1 X 
 
 q= ir 
 
 = 1 — 2 sin 2 x , 
 
 
 sin a; cos a* dx 
 
 
 .-. ie ax \ 
 
 1 1 /"* 2 /* 
 
 = - e ar sin a; cos a: / e ax dx-\ / e a * shr x 
 
 a a J a J 
 
 1 nr • 2 2 „ ' 2 •" j " 
 
 .-. u = - e aT snr# -e ar sm 8 cos a; H r , 
 
 a a* a- # d l 
 
 / 4 \ £ a * sin a:, . 
 
 .-. u[ 1 H — - ) = (a sm x — 2 cos an H — j 
 
 \ a" / a~ ' a s 
 
 Sin X (a sin x — 2 cos a:) 
 .-. |4 a- ^ i j . 
 
 <r + 4 «(<r + 4) 
 
90 EXAMPLES ON THE 
 
 ' dx "" (a -f b cos x) 2 
 
 Assume 
 
 r* d x _ A sin a? p d x 
 
 J (a + b cos x) 2 "" a -f & cos a; a / « + & cos # 
 
 Taking the differential coefficients 
 
 1 A cos x (a + b cos #) -f 6 A sin 2 x 
 
 (a + & cos a?) 2 "" (« + 5 cos x) 2 
 
 B(a + Jcosa;) 
 (a + & cos a?) 2 
 
 .-. 1 = A cos a? (a + b cos a?) + b A sin 2 a? + B (a + £ cos a?) 
 = Aacosa; + A 6 cos 2 a? + A#sin 2 a? + Ba + B&cos# 
 = (A a + B&) cos a; -f A& + Ba. 
 
 Equating like powers of (cos x), 
 
 A« + Bi = 0, .'. A = B 
 
 a 
 
 Ab + Ba= 1 
 b 2 
 
 l(-- + .)-!, 
 
 A = ~ 
 
 B = 
 5 
 
 a 2 -& 2 
 d x 
 
 J (« -J- & cos #) 2 
 
 sin a; a r* d x 
 
 b sin a; a n a x 
 
 ar — b 2 a -\- b cos a: a 2 — b 2 J a + b cos x 
 
 1 r — £ sin a 1 /» da? ") 
 
 — a 2 — 6 2 \a + 6 cos a' ,/ a + 6 cos a; J 
 
INTEGRAL CALCULUS. 91 
 
 (23.) To integrate e™ cos fta\ 
 
 Let j) = cos kx dq=. e ax dx t 
 
 dp = — ft sin kx dx q = — , 
 
 a 
 
 /-.n P 77 ajf cos ft.r ft /» . _ _ 
 
 (1) .*. leP* cos £#a£ = h - / ^sinftxao?. 
 
 J a a J 
 
 /^\ /* • 7 ^ ar sin /car ft /» „ _ 
 
 (2) jef" sin ftx</# = / 6°* cos ft#</#. 
 
 k 
 Multiplying equation (2) by - and substituting in equa- 
 
 a 
 
 tion(l) 
 
 /7 7 /, £ 2 \ e ax coskx ke^sinkx 
 e^ cos #.ra# (14---)= 1 - > 
 \ a- J a a~ 
 
 /. T ^(a cos kx + ftsinft#) 
 e** cos kxdx = — - : — -• 
 a* + k l 
 
 (24.) To integrate e—* 1 * sin ft or, 
 
 ;> = sin kx dq = er° x dx, 
 
 dp = k cos ft * q = — » 
 
 /' . , , e- <u s'mkx ft /* - 
 
 (1) .-. /c'-^sin ft a: aj?= h - / ^"^ r " r cos kx ax. 
 
 J « « c/ 
 
 (2) / <?-*" cos £* 
 
 C** Cos fcg ft /* . _ 
 
 ss / r^sin kx 
 
 a a J 
 
92 EXAMPLES ON THE 
 
 Multiplying equation (2) by f - J and substituting in equa- 
 tion (1) 
 
 /• 
 
 q— ax g j n foxdx 
 
 e -ax s i n fr x k e - ax oos kx 
 
 a ar 
 
 k 2 
 
 e™ sin kx dx 
 
 or J 
 
 cfi -f k 2 r . _ _ er** (a sin kx + k cos kx 
 -— / e -ax sm &# & x ___ v , 
 
 a 1 J a 2 
 
 / mm . , tf-** (« sin kx -f & cos fc#) 
 e~ a * sinhrfa = 5 = n 
 a 4 4- a;" 
 
 (25) f d6 ~ f ***"* 
 
 } J a cos 2 6 + 6 sin 2 ^ a + b tan~ (9 
 
 W(tan0) 1 ~i/ /J\ 
 
 */ (1 - * 2 cos 2 <9)§~ "~ / (1 - «* + ** sin 2 tf 
 
 ■/: 
 
 cos0 .. 
 sin d 
 
 --/- 
 
 ft * (ritftf) 
 
 ^^(aw+rr?) 1 
 
INTEGRAL i ai < i i.us. 
 
 1 
 
 sin 6 
 ~~ (1 _ ^) ^/i -^ + * 2 8 in 2 
 
 ___ sin 
 
 "(1-^)^1 - a « C os 2 
 
 (W.) A cos 2 0) 5 cos d 6 = /(I - 2 sin 2 6f cos (9 d 6. 
 
 Let2sin'<9 = * 2 d# = 
 
 \/2 cos e 
 
 __ (1 - a- 2 )! dx _ (1 - 2a ? 2 + * 4 ) <ta 
 
 dx Sl&dx x A dx 
 
 \/2 du = —7 , + - 
 
 = _ v /l-,-( T +— j + gam-"., 
 
 •. \/Zu = sin -1 a; -f- x \/ 1 
 
 n-i 
 
 ar — sin ■ g 
 8*\ 3 
 
 
94 EXAMPLES ON THE 
 
 4- - sin" 1 (>/2 sin 6), 
 
 o 
 w = y/c^T^ S -^(5 -4 + 4 COS 2 0) 
 
 = ~|~ ( 3 + 2 cos 2 ^) \/cos 8 6 + g-£- sin" 1 ( ^/S sin B). 
 
 (9ft \ C sm 0dQ r sin 6 dS 
 
 J \/sin 2 a — sin 2 6 J *J cos 2 6 — cos 2 a 
 
 __ p d (cos 0) 
 
 ~" "" / / »/i 2 = *~ l0 ^ Vv/COS 2 <9 — COS 2 fl + COS0) 
 
 ^ v cos- — cos 2 a v ' 
 
 = — log (^/sin 2 ;* — sin 2 + costf) -f C. 
 If = u , u = — log (cos a) + C, 
 
 ' * sin d , cos a 
 
 f sin d 
 .'. I = log 
 
 •^ ~ \/sin 3 /% — sin 2 A 
 
 a \/ sin 3 a — sin 2 \/sin 2 a — sin 2 + cos 
 
INTEGRAL CALCULUS. 95 
 
 CHAPTER VI. 
 ON DEFINITE INTEGRALS. 
 
 In the process of differentiation all constant quantities which 
 are merely added to, or subtracted from, those quantities 
 which contain the variable disappear; and, on the contrary, 
 after integration, there may he a constant quantity connected 
 with the integral which we have not in that operation ob- 
 tained. The letter C is therefore added to every integral to 
 represent this quantity, and in order to determine its value 
 we must in the first place find what particular value of the 
 variable makes the integral O ; we thus get two equations, 
 both of which contain C, and between these two equations 
 C may be eliminated. For instance, if we haxe J'S or d 4: 
 = a? + C for the general value of the integral, and the pro- 
 blem indicates that for the particular value x = a, the in- 
 tegral becomes O, then O = a J -f C; by subtracting this from 
 the general value of the integral we get ar 1 — a*, in which the 
 constant C has disappeared ; this latter is called the corrected 
 integral, and is written thus, 
 
 I 3x 2 dx = gfi — a 3 . 
 
 J a 
 
 In this expression the value of the integral commences when 
 x = a, and if we give another value to x, say x = b, then we 
 have fully determined the value of the integral, which is now 
 written 
 
 »5 
 
 84* <*#»&'_ a?. 
 
 %J a 
 
 This is called the definite integral, and is said to be taken 
 between the limits .v = b and x = a : the former is called the 
 superior limit, and the latter the inferior limit, and the ope- 
 ration is called integration between limits. 
 
96 EXAMPLES ON THE 
 
 Iii general f f{x) dx = <p (x) + C. 
 
 In order to determine the definite integral, we must, accord- 
 ing to the nature of the problem proposed, assign the proper 
 limits, which we shall represent by a and b, as before; then 
 we have 
 
 %J a 
 
 f(r)dx = p(a)-<p(6). 
 
 As every function of x may represent the ordinates of a curve 
 whose abscissa is x, it follows that the operation of inte- 
 grating between limits may be applied to finding the areas 
 and lengths of curves, the volumes and surfaces of solids of 
 revolution, &c. 
 
 Examples. — Areas of Curves, Volumes of Solids, Sc. 
 
 (1.) The general equation to a parabola of any order is 
 ym + n _ axm n Then, since A =fydx taken between the 
 proper limits, we have in this case 
 
 A-yv 
 
 ,ni + n „»i+n 
 
 dx 
 
 m + n 
 
 m + 2?i 
 
 a m+n ^m + n + C# 
 
 Then we perceive that the area = when x = 0, .-. C = ; 
 and taking the above between the limits x = and x = x, 
 since the value of the integral commences when x = and 
 ends when x = x, we have 
 
 /.; 
 
 
INTEGRA!, CALCULUS. 97 
 
 (2.) The general equation to hyperbolas referred to their 
 asymptotes is 
 
 m+n 
 
 a m + n , a 
 
 ,\ y n = , and y = — 
 
 9 x m * 
 
 -» ~~n~ r> m+n -- 
 
 •. A = I ydx = / dx = / a n x n dx 
 
 xn 
 
 m+n n— m 
 11 n n . „ 
 
 a x + C. 
 
 We must here determine the constant C, as in the above 
 example, that is, find when the value of x makes the area 
 = 0, &c. We must here observe, that this formula fails 
 when m = n ; for then A = C -f- or log x, which cannot be 
 determined by the above method. 
 
 (3.) The equation to the tractrix is 
 dy y 
 
 and 
 
 dx (a 2 — y 2 )4 
 
 . . ydx = —dy(a 2 — /)*, 
 
 A —Cydx = — J dy {a 2 — y) • 
 
 = ->*-> T-S--5 + C- 
 
 Tl ifii in order to find the area included by the positive axes, 
 let y = — a, observing that C = 0, 
 
 F 
 
98 EXAMPLES ON THE 
 
 whole area = — — , v sin 1 = -• 
 4 2 
 
 Since it is shown in all works on the Differential Calculus 
 that 
 
 dV 
 
 dx 
 
 -.»■- £-..«V> + (£)" 
 
 it follows that to find the volumes and surfaces of solids we 
 have to integrate these functions between the proper limits. 
 
 Examples. 
 
 (1.) To find the volume and surface of a sphere. 
 
 The equation to the circle referred to the centre is 
 y 1 = r 2 — x 2 , where r represents the radius of the sphere ; 
 and as one value of r lies wholly above and the other wholly 
 below the axis of x, we must integrate between the limits 
 x — — r and x = r ; we have 
 
 n + r 4tt 
 
 V = * / {f - x 2 ) dx = ~— r\ 
 
 If we integrated this without reference to limits, the ex- 
 pression we should have would give the volume of the segment 
 of a sphere ; and we observe that C = when x = 0, since 
 the integral becomes 0. 
 
 Also, s = % w f 9 aJi + ( |)V 3*/; vA +. £ 
 
 = 2tt /v/y'^-f- x 2 dx 
 
 = 3*Y V) tftf=aVx 2r 2 = 47rr 2 . 
 
INTEGBAL CALCULUS. 99 
 
 We might form the integral 
 
 ZirfJJf +x l )dx= c l<x f>J7 l dx 
 
 by writing the value of y 2 given in the equation to the curve. 
 Hence the integral = Brra?, which is the surface of the 
 segment, whose height is x. 
 
 ^(2.) To find the volume and surface of a prolate spheroid 
 formed by the revolution of an ellipse about its major diameter. 
 
 b 2 
 The equation to the ellipse is y 2 = — (a 2 — x 2 ), where a 
 
 and b represent the major and minor semi-axes respectively. 
 Hence, V = 77 / y 2 dx = - -j f(a 2 — x 1 ) dx 
 
 which is the volume of a spheroidal segment, remarking that 
 C = 0. Next we must integrate between the limits x = — a 
 and x = a; then we have 
 
 b 2 /*+« 4 
 
 V = tt — / (a 2 - x 2 )dx = - 7:b 2 a, 
 a~ J -a 3 
 
 and — - = — 
 
 \dxj a-(a--x') 
 
 dx a (a 2 — x 2 )* \dxj a 2 (a 2 —x l ) 
 
 a 2 — V 
 
 h we write r for — = — we have 
 ar 
 
 1 _l f*9\*__ * — ** 
 
 f dy\* <r — e 2 x 2 
 
 F 9 
 
100 EXAMPLES ON THE 
 
 nvhec (a 2 \ a 2 . ear) 
 
 a \ \ e 2 J e i a) 
 
 Here C = 0, and if we integrate between the limits x — — a 
 and « = «we have the whole surface 
 
 %irab 
 
 |*(l_a2)* + sm-'e\ : 
 
 (8.) To find the volume and surface of an oblate spheroid. 
 In order to determine the equation to this, we merely have 
 to change a into b, and we have 
 
 and 
 
 
 Integrating between the limits # = — b and a? = -f b, 
 observing that C = 0; for V = when x=0, ,\ the whole 
 volume 
 
 and the surface of the oblate sphere may be found in the 
 same manner as the last. 
 
 (4.) To find the volume of a circular spindle. 
 
 Here let O be the centre of the circle of which -ACB is a 
 segment, and let OC = the rad. = r be perpendicular to 
 AB, OD = a ; GD = x and the chord AB = ^c; GF = y ; 
 (see figure page 103 ; ) then we have 
 
 r 2 = a? + O + a) 2 , 
 
[NTE0BAL 0AL0U1 101 
 
 .-. y 1 = r 2 — a 2 — x 2 — 2ay, 
 
 V = x f'fdx = Tr/^r 2 — a 2 — x* — 2ay) J# 
 
 = r {^J jj «•)* + _ _ Zafydx} 
 
 X s 
 = 7t {(r 2 — a 2 )x — — — 2a x gen. area DGFC} + C. 
 o 
 
 Here C = 0, and the above gives the volume of the frus- 
 tum HECE, the double of which is the whole frustum HFIL. 
 The limits are # = — c, and x = -f c, we have volume of the 
 spindle 
 
 = 27r{Jc 3 — a x gen. area A CB}. 
 
 (5.) To find the volume of an el- Jr^f^^ 
 
 lip tic spindle. 
 
 Let ACB be the generating arc 
 of the ellipse, in which AD = c, / 
 OD = i, MO = a = semi-axis major, 
 and OC = b = semi-axis minor; DG = x, and FG = y. 
 Then from the property of the ellipse c being the 
 
 a : b : : (a 2 - x 2 )* : - (a 2 - xf = PF. 
 
 Hence y = — — — t, 
 
 a 
 
 ••• y 2 = — (a~ — x 2 ) h r, 
 
 .-. V = 7t I y 2 dx 
 
 
 <f« 
 
 = * /{* * - i*_ SW . - (a' - •*)' + 2 i 
 
 3c 2 — # 
 = volume of ECFH, and when x = c, C = o, 
 
 = tt{^ 2 # . — ^ 2* • area DGFC} + C 
 
1.03 
 
 .*. we have it {b 2 c . 
 Zb 2 
 
 EXAMPLES ON THE 
 
 3c 2 -~c' 
 
 3 a 2 
 
 %i . area DAC}, 
 
 or, v < £—3 . c 3 — 2i . area DAct 
 
 the double of which will give the volume of the whole 
 spindle. 
 
 (6.) To find the volume of a hy- £- A 
 perbolic spindle. 
 
 Putting i = the central distance 
 O D, c = AB, and retaining the 
 notation employed in the last, we 
 have by the nature of the curve, 
 
 OM : OC = b : : >/(a 2 + x 1 ) : NF = 
 
 b >/ a 2 + x 
 
 . . V = 
 
 .-. y = OD - NF sa i — 
 = fydx 
 
 b s/a* + tf 
 
 
 ^y^-^ + ^ + s^j^ 
 r^. * 2 */i . * 2 \l 
 
 = *■ J 2t X area GFCD . I - c 2 — — J > + C 
 
 L- ^ & 2 # /3c 2 ~4a? 2 \) 
 = ir |2tx area GFCD- — x f ^ )\ + C ' 
 
INTEGRAL CALCULUS. 
 
 103 
 
 and C = 0. This is the volume of the frustum FCEH, and 
 for the volume of the spindle we have 
 
 Vc 3 
 
 InlSi x areaAFCD — — - 2 j 
 
 (7.) To find the surface of a cir- 
 cular spindle. 
 
 Retaining the notation used in 
 the last, we have 
 
 no = y -f- a 
 
 = n/of' — Yri 1 = \/r 2 — x 2 , 
 
 dy 
 
 x . rdyy x- 
 
 and — = and [ ~ ) = - =» 
 
 ax f^ _ x 2 )* \dx J r l — x l 
 
 = °,7rr / i 1 r ydx = %jrr\x — a sin -1 -f+C, 
 
 and C=0, the surface of the part HECF; and if x=c, we have 
 S = 2 7rrjc — « sin -1 — >, 
 
 the double of which will be the surface of the whole spindle 
 ACBE 
 
104 
 
 EXAMPLES ON THE 
 
 (8.) To find the volume of a 
 parabolic spindle. 
 
 Put CD = h, AB = 2c, and A 
 
 # = AG, and y = FG. 
 
 From the property of the parabola 
 
 AD 2 : AG.GB :: CD : EG, 
 
 c 1 : x (2 c — x) : : h : y, 
 
 h(%cx — x 2 ) 9 h 2 2 . 3 4 . 
 
 y= — 2 -, .'. f = — (4c 2 # 2 — lex* + #), 
 
 c 
 
 \ IB / 
 
 = tt— /(4c 2 # 2 — 4ctfr 3 + x 4 )dx=z — — f 
 
 irtf/Ua? 
 
 + +*} 
 
 the volume of AFH, since C = ; when # = c, we have 
 tt/* 2 /4c 5 _ c 5 \ 8 72 
 
 volume of the semi-spindle ; and, as we have found the volume 
 of the part AFH, if we subtract this from the whole semi- 
 spindle, we shall have the frustum EHCF, the double of 
 which will be volume of the whole frustum EHIL. 
 
 In the same manner as in the last example, we may find 
 the surface of the parabolic spindle. 
 
 (9.) In a parabola find the area included between the 
 curve, its evolute, and its radius of 
 curvature. 
 
 area parabola ANP = Cy dx 
 
 4 2 
 
 = 2 s/a I s/x dx = - \/ax' 1 t 
 J 3 
 
 area evolute A N 
 
 
INTEGRAL CALCULUS. 105 
 
 (a _ 2a)* = i— 9 jsJ = l 
 
 15^/8 a i:. v/3« B ^ a 
 
 dy 2a 
 
 subnormal NG = y — - = y — = 2 a, 
 o y 
 
 NP . NO I'i 1 
 
 area PNG = = ay = 2 a # 
 
 ON' = AN'- AG = 34? -f 2a — a? — 2a = 2#, 
 
 GN'. N'P' . y* 
 
 area P'N'G = = fa = -f-, a? 
 
 2 4a- 
 
 4a# 2 h v 2a; 5 
 
 = r- 2 v/^^ = ' 
 
 4a- \/a 
 
 area APP' = APN + NPG + AN'P'- GN'P' 
 
 5 
 
 4 | J i 12# 5 
 
 = -,/„* + 2a^ + ^ — - — 
 
 3 l 5 
 
 20a# 5 + 30« 2 * a + 6a; 5 
 
 15 v^a 
 
 = [ cr + - ax + -x- . 
 
 -v/a \ 3 5 / 
 
 (10.) To find the length of the spiral of Archimedes. 
 
 dr 
 
 ds d$ dti 1 a-5 z 
 
 T = TH ' T" *" " V r ~ + a 
 
 F 8 
 
106 EXAMPLES ON THE 
 
 — \f rjr — 2 a Y r r2dr , f dr 
 
 a J yV + a 2 J sj r l + a 1 
 
 = r v 7 ** 2 + « 2 — /\*r \A 2 + «* + « r , r - 
 
 a *J J sjr 1 + a 2 
 
 r /-= a r dr n dr 
 
 2 « ^ */ v/r 8 + a 2 ./ v/r 2 + a 2 
 
 2« v 2 7 v/r'-H-a 2 
 
 ' /H^ 2 . a 1 f V^ 2 + « 2 + r \ 
 
 (11.) On AB, the diameter of a given semicircle ACB, take 
 AD = the chord AC; join C, D; bisect C D in P, and find 
 the equation and area of the locus of P. 
 
 Join A, P, and put A P = y, /_ PAD = <p, and AB = 2 a; 
 then AC =AD = 2« cos 2 $, and A P or y = 2 a cos £> cos 2 p. 
 Let A P = y be drawn in an opposite direction ; then 
 <p will become 180° -f <J>, and A P or y will become = 
 2acos(180°=<p)xcos(360°-f 2<£)= — 2acospcos2<£ = — y, 
 which indicates that when $ passes 180°, 
 the point P will begin at B, and de- 
 scribe exactly the same curve which it 
 has described ; hence, when <p arrives at 
 180°, the curve is complete. When 
 y = 0, $ has three values from 0° to 
 180°, viz., 45°, 90°, and 135°, which shows that the point P 
 returns thrice to the point A, and therefore describes three 
 noduses; 2 area ABP = fy 2 d <p = ka 2 fd <p cos 2 <p (1 — 4 sin 2 (p 
 cos 2 $) = a 2 <p -f -£- a 2 sin <p (3 cos <p — 2 cos 3 (p -f- 8 cos 5 #), 
 this between <p = 0°, and <p = 45°, gives 5 circle + § a 2 , or 
 1-45206 a 2 for the area APBEA, and between (p = 45°, and 
 (?) = 90°, gives i circle — § a 2 , or -118733' a 2 for the area 
 of the other two small and equal noduses, each of which 
 are therefore = | circle — ^ a 2 , or -05936' a 2 ; hence the 
 area of the entire curve comprehending the three noduses 
 is equal to the semicircle ACB. 
 
INTEGRAL CALCULUS. 
 
 107 
 
 (12.) ACDB is a given circle, whose diameter is A B ; on 
 the chord A D, which is an arithmetical mean between the 
 chord A C and the diameter A B, take A P, a 
 geometrical mean between A C and A B, 
 -linw the nature and area of the curve 
 which is the locus of P. 
 
 Put AB = a, AP=r, and Z. PAB 
 = 9 ; then A D = a cos G, and AC=2AD 
 — A B = 2 a cos 6 — a, and consequently 
 AP 2 or r 2 = AC . AB = a~ (2 cos — 1), 
 the polar equation to the curve. For the quadrature we have 
 
 2areaABP = fr>d6 = a 2 fdS (2 cos 6 — 1) 
 
 = 2 a 2 sin 6 — a 2 6. 
 
 To ascertain the limit, put r = 0; then r 2 or a 2 (2 cos 9—1) 
 = 0, and cos 6 = % ; therefore the curve makes an angle of 
 60° with the diameter A B ; hence, taking the above between 
 
 9 = 0, and 6 = 60°, we get a 2 \/3 — - a 2 . 2tt, or a 2 x '6848. 
 
 (13.) Let the given circle CED, whose diameter is CD, 
 touch the indefinite right line A D in D, and from the given 
 point A, draw the right line A EC, 
 on which take AP = the sine of the c 
 
 arc EC, and find the equation and 
 area of the curve which is the 
 locus of P. 
 
 The lines being drawn as enun- 
 ciated, put CD = a, /_ DAC = <b ; 
 then AD = acotp, and, by similar 
 As, CD : AD : : CG : EG =cotp x CG; but, by the circle, 
 
 A . a 
 
 E G = (a . C G — C G*) ; whence CG = — = a 
 
 v ' cot 2 <p + 1 
 
 sin 2 <p, E G = A P = (a 2 sin 3 <p — a* sin 4 of = ^ a sin 20; 
 the polar equation of the curve. When = 90°, AP = (), 
 when <p = 45°, A P = £ a, and when $ = 0°, A P = ; hence 
 A is a punctum duplex. 
 
108 EXAMPLES ON THE 
 
 Quadrature. — f\ AP x dtp = -J a 2 C d(p sin 2 2 (p = J a 2 
 
 (— | sin 4 <p -f ^p), which, between <p = 90°, and (p = 0°, 
 gives J of the given circle for the area of the nodus A PA. 
 
 (12.) Let AB, the base of the right-angled triangle ABC, 
 be produced till B P be equal to the diameter of the inscribed 
 circle. Required the equation, quadrature, and 
 greatest possible ordinate of the curve which is ^ — x 
 the locus of P, the hypothenuse AC being given, f/£v/\ 
 
 Put the hypothenuse A C =a, 3*14159265 =tt, VZ__^ 
 and angle CAB = a;; then, by trigonometry, A B c a 
 
 = a cos x, and B C = a sin x ; whence B P = 
 AB + BC — AC = a (cos x -f sin x — 1), and A P = a 
 (2 cos x -f sin x — 1), the polar equation of the curve. When 
 x = 90°, AP = 0, and when x = 0°, AP=a; .*. the curve 
 commences at A and terminates at C. The maximum polar 
 ordinate is when cos x = 2 sin x, or when sin x = 4- ^5, 
 or x = 26° 33' 9". 
 
 Quadrature, — The differential of the area is = £ AP 2 dx 
 = ^ a 2 dx x (4 cos 2 x -j- 4 cos x sin x -j- sin 2 x — 4 cos x — 2 
 
 / 7 3 
 
 sin x -j- 1) = \ a 2 dx X ( - — 4 cos x — 2 sin x -j- - cos 2 x 
 
 7 x 
 2~ 
 
 -f- 2 sin 2 x J, and the integrals give the area = J a 2 x ( 
 
 3 \ 
 
 — 4 sin # + 2 cos x + - sin 2 # — cos 2 x J ; which, be 
 
 tween x — 0°, and # = 90°, gives § a 2 n — 2 a 2 for the area 
 of the whole curve A PC, or in numbers = '74889357 a 2 , as 
 required. 
 
 The area of the space inclosed by the curve 
 
 ay 1 = x 1 v a 1 — x 2 is - a 2 , 
 o 
 
 Area = j ydx = — - / x (a 2 — x 2 )* dx 
 
 = -h-\^-*f +c 
 
I\ I EGRAL CALCULUS. 100 
 
 If x = () Area = - dr + C. 
 
 5 
 
 If x = a Area = 0, 
 
 4 . 2 
 .*. Whole area = or. 
 
 The length of the curve, y log ( — J 
 
 from x = 1 to # = 2 is = log (e ■+- -1 ). 
 
 <fy __ «* (e* — 1) — ** (e* + 1) e* — 1 
 
 d# " (e r — I)- ' e + 1 
 
 — 2^ 
 " «-* — 1 ' 
 
 J/ ^ g4j _ 2^ + 1 + 4e 2 * _/e 2x +lY 2 
 + dx^~~~ (e 2 * - l) 2 "" U> 2 *- 1A 
 
 - -/< V^g-y^f* 
 
 r e* dx r d x r e x dx r e x dx r 
 
 m J ?TT * J i 7 ^! "V ^TT + ./ JTZT "V ■ 
 
 S = log(<r x — i) — * + C. 
 If a? = 2 S = log (<? 4 - 1) — 2 + C. 
 
 *«. ] S = log(* 2 — 1) - 1 + C, 
 
110 EXAMPLES ON THE 
 
 .-. from* = 2 to x = l S = log(> 4 -- 1) — log(<? 2 — 1) — J, 
 S=log(* 2 + l)-log* 
 
 —j- J = log (« + 0- 1 ). 
 
 (14.) The length of the curve 8a :3 y = x x 4- 6 a 2 a? 2 , 
 measured from the origin of co-ordinates, is 
 
 h ( *' + 4a5)l 
 
 1 + -r£ = A (*» + 6a 2 * 4 + 9a'*' + 4a 6 ) 
 rfj»' 4a° 
 
 = — {a* + 3a 2 * 4 + 3a 4 * 2 + a 8 + 3<r (* 4 + 2 a* a? 8 + a 4 )} 
 
 4 a 8 
 
 = j^ j(* 2 + a 2 ) 3 + 3a 2 (* 2 + a 2 ) 2 j 
 
 (* 2 + a 2 ) 2 
 
 4a , (** + 4a*), 
 
 ■■■>=J d 'V i+ %=^j(*' ! +«w* 
 
 ~\- ±a 2 dx 
 1 # * 
 
INTEGRAL CALCULUS. 1 1 I 
 
 (15.) The volume generated by the curve y 2 (x — 4 a) = 
 a x (x — 3 a) revolving about the axis of x, from x = <> to 
 
 j: = 3a is = -*ra 3 (15 — 16 log 2). 
 
 # 2 — 3a# / 4a 2 \ 
 
 x — 4a V # — 4a/ 
 
 .-. V = 7T I f dx = ira (- + a# + 4 a 2 log (x — 4a) J + C. 
 If * = = 4*ra 3 log (- 4a) + C. 
 
 If x = 3 a 
 
 / Oa 2 \ 
 
 = n a I — + 3 a 2 + 4 a 2 log ( - a) J + C, 
 
 '• v = na "TT" "" 4a2 { log ( ~~ 4 «) - lQ g(- «)}} 
 
 = ^ a 3Jl5-81og4} 
 
 = ^a 3 |l5 - 16 log iij. 
 
 (16.) Find the area of the curve in which 
 (a 2 - b 2 ) sin 9 cos 9 
 
 T = — — — 
 
 s/a 2 siir 6 + 6 3 cos 2 
 
 1 r^it l r( a2 - 62 )* 2 sin " e cos 8 6 _ . 
 
 area = - / r 2 dQ = - / v „ . — ; ( / 
 
 2 % / 2 J a 2 siir 6 + V cos 3 
 
 _ l Wq 2 -6 2 ) 2 si n 2 fld0 
 ~2j a 2 tan 2 5"+ b 2 * 
 
112 EXAMPLES ON THE 
 
 1 . X 
 
 Let x = tan 6, d6 = dx sin 6 = 
 
 1 + ^ ?2 \A+^ 
 
 area = - / * ' 1 + a? 2 ] + x 2 
 
 J a 2 x 2 + b 2 
 
 I r {a 2 - b 2 ) 2 x 2 dx 
 ~%J (a 2 x 2 + b 2 )(l + *T 
 
 U A B C 
 
 Let -| = TXT" To + /, ■ -2N2 + 
 
 V " a 2 # 2 + & 2 ' (1 + x 2 ) 2 1 + «* 
 = A(l + x 2 ) 2 + B (> 2 # 2 + 5 2 ) + c (1 + **) (a 2 x 2 + 6 2 ). 
 
 A 2 Z> 2 / a 2 — h 2 \ 2 
 
 Let rf = - L .'. - (a 2 - 6*)*1 = A f ^__^) 
 
 A = - a 2 5 2 . 
 
 Let a = v /:r " 1 5 - (<* 2 - **) 2 = - B {a 2 - 6 2 ), 
 .-. B = a 2 - £ 2 , 
 
 ... ( a 2 _ £2)2 tf _ ( a 2 _ £2) (a 2 ^2 + £2) + tf £2 (J + ^ 
 
 = C(1+^ (a 2 2 + 6 2 ) 
 
 _ ( a * _ £2) (] + a-') b* + a 2 5 a (1 4- xj = C(l + x 7 -) 
 
 (a 2 2 + £ 2 ) 
 
 b 2 (1 + 2 ) (« 2 +a 2 ^-a 3 + ^) = C(l+ 2 ) (a 2 rr 2 -f- & 2 ) 
 C = b\ 
 
RAX CALCULUS. I 18 
 
 r> {a* —b ) x" dx 
 'V (a 2 a* + fc 2 ) (1 + *<)* 
 
 p dx p dx p x*dx 
 
 4 J (i + jy ~J Hh? ~7 (i + **)* 
 
 /• rf# « 1 /* </# 
 
 1 +x> + 2 (1 + x 2 ) ~ 2./ l+* ! 
 
 1 p dx 
 = 2 7 IT"? 
 
 d x m x 
 
 2~(r+ *7 
 
 a a fc 2 r_&x a 2 + & 2 /" </*? 
 
 •'• area " " TV a*** +6* + — I~J IT* 
 
 + 
 
 4(1 +*') 
 
 d b . a x a 1 + b z 
 
 = tan"" 1 — + — -7 — 
 
 tan -1 x -f- 
 
 9 b ' 4 4 (1 + **) 
 
 Taking this between limits = and = 90 ; 
 
 or, # = x = oo 
 
 a 2 + /; 2 7r a 5 7T (« — £>) 2 r 
 
 area = - — = - - -. 
 
 4 2 2 2 4 2 
 
114 EXAMPLES ON THE 
 
 (17.) The length of the epicycloid after one revolution of 
 the generating circle = 8 - (a + b), and the area between 
 
 the epicycloid and the circle = ?r b z ( 3 -\ ) 
 
 ( r 2 fl 2x 
 — 2 g )» equation to epicycloid in terms of rad. 
 
 vector and perp. on tangent where c = a -f 2 b. 
 
 ds r 
 
 dr ~~ yV _ f 
 
 , J; cV~aV- c V + flV _ a 2 (c 2 - r 2 ) 
 
 r 2 — ^) 
 
 e?s n/c 2 — a 2 r 
 
 n/c 2 — a 2 , 
 
 * = ± n/c 2 - r 2 + C. 
 
 If r = a + 25 = c, then s =: 0, C = 0. 
 
 c 2 -« 2 «* + 4a6 + 4** — a 3 
 
 If r = a, then s = • 
 
 « = ± — (a + b), 
 
 hence whole length of arc of epicycloid 
 
INTEGRAL CALCULUS. 1 1 5 
 
 d6 V 
 
 Also to find area — = ,_ 
 
 M 2 
 
 *f' r>/r 2 —p 2 
 
 , , - prdr cr \/ r 2 — a 2 Wc 2 — a 2 aV 
 
 rd$= — = — v X — =^ 
 
 ^/r* __ ^ y^ _ a * a ,/c 2 - r 2 
 
 c r^/ r 2 — a 2 dr 
 
 1 /*«jii c rrdr\Jr i — d i 
 area =- / r 2 d0 = — / v 
 
 Let f* — O 1 « **, raVssfcda?, c 2 — r 2 = c 2 — a 2 — ;*r, 
 
 c r z l dz 
 
 z 2 dz 
 
 area = — -r-^v/S® — ^®+ -^ — sin" 1 — -f C. 
 4a 4a £ 
 
 If r = a then z = 0, and area = 0, 
 r = e then # a — *» = z* = c 2 — a', 
 
116 EXAMPLES ON THE 
 
 c(c 2 -a 2 ) . , \A 2 - « 2 
 .-. semi-area = ~— t sm -1 • /" ■ ■ 
 
 4 « VV-a 2 
 
 __ c (c 2 — a 2 ) ?r 
 ~ la 2 
 
 area circle = - %>n-b =z nab, 
 
 A 
 
 c ( c i a i\ 
 
 area between epicycloid and circle = — -n — mab 
 
 4a 
 
 (a + Zb)±b(a + b) 
 
 7T — nab 
 
 4a 
 
 = -(a 2 5 + 3a5 + 2P — a'b) 
 
 (18.) Find the length of the curve where 
 
 fl?t -(- yt s-s a t , 
 
 - p\/ "*+'' =f- --<•> 
 
 Taking* it between the limits x = 0, # = a, 
 
 3 
 
 s = -a. 
 
 The whole length of the curve 4 x - a = 6 a. 
 
 x* ■ 
 
INTEGRAL CALCULUS. 117 
 
 (19.) If h as height of a parabolic frustum, a and b the 
 radii of the ends, show that 
 
 Frustum = ~ (a 2 + b 2 ). 
 
 A 
 
 Equation to parabola y 1 = kmx. 
 
 / I + h / x + h 
 
 V = ?r/ y 2 dx as kitm I xdx 
 
 = 2 ttn {{x + hf — x°] = 2 %m (2 ar/i + A 2 ) 
 
 (20.) Find the area of the catenary 
 
 Area = j yd* = \ f [f + e" a ) dx 
 
 at - -*\ 
 
 = - ( ae a — ae * h 
 
 / 2 JL _2j 
 
 = - y/ty* — 4a 2 = a \J y' — a' 2 - 
 
 (31.) Find the area of #V — a V = " s > 
 
 a 2 
 
 y = «/-« 4 ' 
 
 sj x* — a 4 
 
1 I 8 EXAMPLES ON THE 
 
 area= / ydx = a 2 / , — = a 2 I — , 
 
 J J {/a;* -a* J ** 
 
 If a? 4 - a 4 = *V, 
 
 VT=? 
 
 log x = log a — - log (1 — z% 
 
 dx Z* 
 
 x 1 — z* 
 
 .-. area = a 2 I - - r dz = a 2 / - dz 
 
 J z \ — sr J \ ~ sr 
 
 2 (J l-z 2 J 1 + z 2 S 
 _ a 2 r 1 a 2 r 1 fl /' ] 
 
 a 2 _ 1 + * a 2 
 
 = 7 log r3 
 
 * 4 - a 4 a 3 
 
 tan" 1 #. 
 
 But* 4 =- — , 
 
 a; 4 ary 4 
 
 a 2 xy a~ a 2 
 
 area = T log 1 — — tan — 
 
 = ^!{l 0gA /fl^!_tan-^}. 
 
INTEGRAL CALCULUS. 110 
 
 (22.) Find the volume generated by the revolution of the 
 witch round its asymptote. 
 
 Qa — x 
 y 1 = 4 a 2 equation to witch, 
 
 xy* = 8 a 3 — 4 a 2 a?, 
 
 8 a 3 
 * - y 2 + 4 a 2 ' 
 
 /* /* 4a 2 a , y 
 
 * 2 dy = lGa V (y * + 4a 2 ) 2 
 
 ./ i/ 2 + 4a 2 ./y + 4^)F 
 
 B t /■ y dy_ _ 1 y . I / 
 
 \/ ^ ( y » + 4a a)« 2 y 2 + 4a 2 "*" 2 ./ 
 
 dy 
 
 y 2 + 4a- 
 
 Volume V 
 
 = 8a 4 9r 2 , y - 2 + 4=a 3 7r tan" 1 £-. 
 
 Taking between the limits of 2/ = oo and y = 0. 
 
 Volume = 4a 3 7r tan -1 00 = 2 a 3 tt*, and whole volume 
 generated by curve both above and below the abscissa 
 = 4 a" «* 
 
120 EXAMPLES ON THE 
 
 MISCELLANEOUS EXAMPLES. 
 
 C 1 -) fs/bax — x 2 . dx 
 
 /{lax — x 2 )dx 
 s/%ax — x 2 
 
 ~xdx {a + (a — x)} 
 
 -j. 
 
 sj %ax — x 2 
 
 (a — x) xdx 
 
 r* axdx p\a — x) xdx 
 
 J \Z~^ax — x 2 J s/%ax — x 2 
 
 /(a — x) dx . x 
 \/ %ax — x 2 
 
 = x \/%ax — x 2 — J \/%ax — x 2 dx, 
 
 /► axdx P{( a — x ) a ~ <^}dx 
 
 sj^ax—x 2 J \Z%ax — x 2 
 
 = — a \/^ax — # 2 + a 2 . ver sin -1 -, 
 
 .-. 2 J \/%ax — x 2 . dx 
 = (# — a) y/% ax — x 2 + a 2 ver sin -4 - , 
 
INTEGRAL CALCULU8. 1-21 
 
 .% J \Z%aw — x l . dx 
 
 (x — a)\/2ax — x l a 2 . ,m 
 
 ; Fi + 77 ver 8in • 
 
 2 2 a 
 
 This integral is used by Earnshaw, in finding the centre of 
 gravity of the area of a cycloid. 
 
 /§f d x 
 >/&** + </ 
 
 By for mula of Reduction, 
 
 /x" dx 
 s/Zax + a 2 
 
 
 x*" 1 \/ %a x + x l a(2n— 1) /» x n ~ l dx 
 
 \'%a+ x"- 
 
 or dx 
 
 If ?i = 2, this gives / . 
 
 J s/^Xax + *■ 
 
 #v/2a# + ar 6a n aria 
 
 2 
 Also /» "'« 
 
 6a p x ax 
 2 •/ s/^ax -f-jr 1 
 
 /* d x 
 
 o 
 
122 EXAMPLES ON THE 
 
 And d * 
 
 f* , : = log {x + a + s/%ax + ar}, 
 
 «/ \/%ax-\- x 2 
 ss >/2 a # -j- a? 2 — a log {a + a -f \/2ax 4- # 2 }. 
 
 z^ a? 2 ^ #\/2a# + # 2 3a /- - 
 
 ,\ / — , = — - \/%ax -{- x 2 
 
 J ^Zax + X 2 2 2 
 
 g^2 
 
 + — - log {x + a + ^/2a# -f # 2 }. 
 
 /» ° .i? 2 a 7 ;*? 
 
 let ^? = x dp = cfar, a 7 # = 
 
 (2a# — a; 2 )i 
 x dx 
 
 /x dx /v 3 i 
 = / (2a — a?) _ 2 0* d# 3- 
 
 / (2a# tal — l) - '3 *"* ^ 
 == - -i- f(Zax- 1 - 1)~~* x — 2a x" 2 dx 
 
 ° ' «V 2a-V 
 
INTEGRAL CALCULUS. 123 
 
 j pdq=pg -fqdp 
 
 _ x / x - C sjx ' dx 
 
 ~ a v 2a — x a J sJ'Za — x 
 
 _ x / x i r xdx . 
 
 ~a\ %a — x a J \/%ax — zr 
 but by the formula of reduction, 
 
 r x dx /- r , x 
 
 I — = — w %ax — x* + a vers ' -, 
 
 /x 2 dx 
 (tax - x*f 
 
 X / X 1 / 
 
 x — j? 8 — vers -1 -, 
 a 
 
 taken between the limits of and a, 
 x 2 dx 
 
 /. 
 
 ; sl + l-! = 2-!= -4292. 
 
 a (2ax - z 2 )i 2 2 
 
 This integral is used in Barlow on the strength of mate 
 rials. See pages 364, 365. 
 
 x n x*dx / 1 x — a 1 ,#\1 
 
 /» rfx __ x 1 1 «JL 
 
 (5 '!/ W+W -*h*? + i>) + 2*7i tan v ^' 
 
 o 2 
 
121. EXAMPLES ON THE 
 
 /* dx 
 
 (x -t a) (** + bf 
 
 1 , x + a 1 ax + b 
 
 a (a 2 4- 36) , x 
 -f ^ 7 - , - — iT7, 'tan -1 -, 
 
 x 4 dx 
 
 r x* dx 
 
 (7,) J (« + bxy 
 
 /$a? Sax\ 1 3 /• 
 
 ~ "" { ~8b + "8F/ ' O + bxf+ Sb 2 J a 
 
 dx 
 
 -h bx 2 
 
 L' {a -r bxf 6 3V o 3 « -f &# 6* 7 
 
 The formula of reduction 
 
 dx _ ^Jx—x 1 
 
 (9 .) /' ^ = = 
 
 — A 
 
 
 W -V ^— 2 v' 1 - OJ 2 
 gives the following integrals, taking m odd 
 
 ydfa? \/ 1 — x l 1 / * dx 
 
 y<2# sj 1 — a; 2 3 / • </a; 
 
INTEGRAL CALCULUS. L96 
 
 / dx y/l — ar 5 r dx 
 
 J x 7 x/T^x 1 " 6 *" f 6 c / ^ N /T37' 
 
 Taking m even 
 
 cto v/l— ^ 2 /* dx 
 
 
 / ' </# _ v/ 1 — x' 2 r ax 
 
 &c. 
 The following integrals may be done by the general form 
 
 / x"' 1 (a -f- bx n )'i . 
 
 ydx 1 . a 4- 6# — y/a 
 
 #\/# + £# %/# a 4- £# 4- v# 
 
 /» </# k/± + Sx 3, \/4 4-3# — 2 
 
 /i i \ / === = — log v — • 
 
 ^ %9 J xs/T~+Vx ±*> ™ v/4 + 3^ + 2 
 
 ydx 2 . _ /&# — a 
 
 f dx 2 
 
 (13.) / 5= 7== 
 
 * 7 a? (a + &#)* a v 7 a 4- i 
 
 >\Jbx — a \J i 
 dx 
 
 ~bx 
 
 log 
 
 a \/ s/a + bx + s/a 
 
 r x Q dx J f4 - 4# 4- 5a?*| 
 
 y» d# _ / 1 2 \ a? 
 
 (a + fc* 2 )* " \ 3a(a + ft**) + 3^ J ^JT*"£? ' 
 
 J* X\dx iM f 1 .2 1 
 
126 EXAMPLES ON THE 
 
 /__d®_ f 2 2_~l 
 
 a? + a 
 
 <tf \J % ax -f # 2 
 
 ne x xdx e x 
 
 (20.) Show that * J 8in * = tan o 
 (31.) feS x xdx = Ze^ x (x*-3x + 6 ^/a — 6). 
 aJ? sin 2 #6?# 
 
 a (a 2 + 4) 
 
 — ; — (a sin # + 2 cos #), 
 
 a 2 + 4 
 
 /#* <^costf# rsm^ 
 e , cog8a! = _ + ___ + ___. 
 
 (24 .) /L_^_ = 2tan, +^ + -i- 
 v y ,/sm 2 tfcos 4 # 3 tmx 
 
 also, show that the integral is 
 
 1 1_ _8 
 
 3 sin x cos 3 x 3 
 
 cot2#*. 
 
 * It is observed by De Morgan, that we are liable, in using artifices of 
 integration to produce results which appear different, but which in fact only 
 differ by a constant. This discrepancy does not appear when the integrals 
 are taken between definite limits, since <p a — <p h = a + C — (<pb + G), 
 are the same. See De Morgan's Differential and Integral Calculus, p. 116. 
 
INTEGRAL CALCULUS. 127 
 
 r dS l j y/?^l coal — sin | 
 
 C20.)y l _ e 2 C08 2 e - 2n / x __ f2 lQ g (/— i cos B + sind J 
 
 n/5 
 
 (« > i). 
 
 /* dd 3 f«0 + 0) 
 
 sin 3 (a + £) cos 2 (a + 0) = 2^ l0g ^ I 2 J 
 
 If 1 cos a 6 + b ^ 
 
 a {cos (a Q + 5) "" 2 (sin 2 a + b)\ 
 
 d B vers 2 
 
 sec cosec 6 
 
 (29.) fd 6 cosec 2 0= - log tan 0, 
 
 r d6 1 
 
 (30.) / — 2 — - = -tann0, 
 
 v ' J cos 2 n 6 n 
 
 ,„ x /* d * * 1 i 1 + sin 2 <9 
 
 , r de i i i, { * Q\ 
 
 ; 8a > ./sto* = suf* - e~# + a log H 4o + a ; 
 
 /^sin 3 d 
 (33.) / - ,, A = cos + sec 0, 
 
 cos 2 
 
 (34.) A— ^ T"/, = « i"i + lo g tan ^ 
 
 v ' ^ sin cos 3 2 cos 2 6 ° 
 
 /^sin 3 6 d 6 _ sin 2 2^ 
 
 ^ ■' J cos 8 ~~ 5 cos 7 5 • 7 cos 7 
 
 (36.) /" / r 2 sin0d0rfr=- -(R 3 - r 3 ) (cos - cos 0). 
 
 This integral is used by Professor Moseley on the arch. 
 See the " Mechanical Principles of Engineering and Archi- 
 tecture," page 467. 
 
 (37.) AOB is a quadrant of a circle, O its centre ; draw any 
 chord BED, and OE perpendicular to it, upon which tak< 
 
128 EXAMPLES ON THE INTEGRAL CALCULUS. 
 
 OP equal to half the cosine of the arc BD, and determine 
 the quadrature of the locus of P. 
 
 (38.) CP and CD are conjugate diameters of an ellipse, of 
 which the semiaxes are a and b ; P F is perpendicular to DC : it 
 is required to find the area of the curve, which is the locus of F. 
 
 (39.) A right line drawn from a given point cuts a given 
 circle, and the intercepted chord is the minor principal axis 
 of an ellipse, whose area is equal to that of the given circle. 
 Find the quadrature of the curve described by the vertex 
 of the ellipse. 
 
 (40.) Supposing the arc of a semicircle to be stretched out 
 into a straight line, and an indefinite number of perpendiculars 
 erected on it, each equal to the versed line of the correspond- 
 ing arc ; what would be the length of the curve traced out by 
 the tops of the perpendiculars ? 
 
 (41.) The polar equation to the parabola is r= - . show 
 
 2 * ; 
 cos 7j 
 
 that the area = a 2 ( tan - + - tan 3 - j. 
 
 (42.) The equation to the lemniscate being (x 2 + y 2 ) 2 = x 2 
 — y\ find its area between the limits of x = and x = 1. 
 
 (43.) Let the base AB of a right-angled plane triangle be 
 given, and in the variable hypothenuse AC, let there be con- 
 tinually taken CP equal to the perpendicular CB. Find the 
 equation and quadrature of the curve, which is the locus of 
 the point P. 
 
 (44.) ACB is a given quadrant of a circle ; A the centre, 
 and D any point in the curve. Draw O D perpendicular to A B, 
 and take D P = B O ; then the area of the curve, which is 
 the locus of P, will be = the circular segment CDBC. 
 
 (45.) The perpendicular BC of aright-angled triangle ABC 
 is given, and in the variable hypothenuse AC, let AP be 
 taken, so as always to be equal to B C ; required the distance 
 BP of nearest approach, and the quadrature of the curve, 
 which is the locus of P. 
 
 (46.) Find the content of the solid generated by the revo- 
 lution of the curve, whose equation is (a 1 -f 0) JJ a — ** 
 ( a 2 — x 2 ) = ; about the axis of x. 
 
 G. Woodfall and Son, Printers, Angel Court, Skinner Street, London. 
 
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