thomasET_226348_ism41

thomasET_226348_ism41 - Chapter 8 Practice Exercises 89-92....

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Unformatted text preview: Chapter 8 Practice Exercises 89-92. Example CAS commands: Mathematica: (functions and domains may vary) Clear[x, f, p] f[x_]:= xp Log[Abs[x]] int = Integrate[f[x], {x, e, 100)] int /. p 2.5 In order to plot the function, a value for p must be selected. p = 3; Plot[f[x], {x, 2.72, 10}] CHAPTER 8 PRACTICE EXERCISES # ' x4x# c 9 dx; " u oe 4x c 9 du oe 8x dx " 8 555 1. ' u du oe " 2 u$# b C oe 8 3 " 1# a4x# c 9b $# bC 2. # ' 6x3x# b 5 dx; " u oe 3x b 5 ' u du oe 2 u$# b C oe 2 a3x# b 5b$# b C 3 3 du oe 6x dx 3. ' x(2x b 1)"# dx; " u oe 2x b 1 du oe 2 dx #$ #& " # ' ^ u c 1 u du oe " S' u$# du c ' u"# du< oe " ^ 2 u&# c 2 u$# b C 4 4 5 3 # oe 4. (2x b 1) 10 c (2x b 1) 6 bC (1 c u) u ' x oe 2 3 1cx dx; " uoe1cx c' du oe c dx du oe ' Su c " u < du oe 2 3 u$# c 2u"# b C (1 c x)$# c 2(1 c x)"# b C ; b1 " # 11. &$ b ' z#$ ^z&$ b 1#$ dz; " u oe z5 #$ " du oe z dz 3 # % 10. ' 2t dt t b1 ;" % 9. ;" u oe t# du oe 2t dt ' u du 1 oe tanc" u b C oe tanc" t# b C b 3 5 ' u#$ du oe 3 3 u&$ b C oe 5 5 % $ ' t dt 9 c 4t % 8. ' 4ybdy ; " y $ # 7. dy ' 25yb y ;" # 6. 9 c 4x ;" u oe 25 b y# du oe 2y dy " 4 " # " " ' du oe # ln kuk b C oe # ln a25 b y# b b C u u oe 4 b y% du oe 4y$ dy " " ' du oe 4 ln kuk b C oe 4 ln a4 b y% b b C u u oe 9 c 4t% " c 16 ' du oe c16t$ dt du u " oe c 16 2u"# b C oe c 9 c 4t 8 # ' x dx # 5. ' x dx 8x u oe 8x# b 1 du oe 16x dx " 16 du " ' u oe 16 2u"# b C oe 8x8 b 1 b C u oe 9 c 4x# c "' 8 du oe c8x dx du u " oe c 8 2u"# b C oe c 9 c 4x 4 bC bC 9 #5 ^z&$ b 1&$ b C 556 12. Chapter 8 Techniques of Integration 5 4 5 ' uc"# du oe 5 2u b C oe # ^1 b z%& "# b C 4 %& 1 ' zc"& ^1 b z%& c"# dz; " u oe 4 b z c"& du oe z dz 5 2) ' (1sincos d))) c 2 15. sin dt ' 3 b 4tcos t ; " u oe 3 b 4 cos t du oe c4 sin t dt " c4' 16. 2t ' 1cos sindt ; " b 2t u oe 1 b sin 2t du oe 2 cos 2t dt " 2 " " ' du oe 2 ln kuk b C oe 2 ln k1 b sin 2tk b C u 17. ' (sin 2x) ecos 2x dx; " u oe cos 2x c " ' eu du oe c " eu b C oe c " ecos 2x b C # # # du oe c2 sin 2x dx u oe sec x ' (sec x tan x) esec x dx; " ' eu du oe eu b C oe esec x b C du oe sec x tan x dx 18. 23. ' v dv v ; " ln 24. dv ' v(2 b ln v) ; " u oe 2 b ln v " # # # 28. ' dx 49cx oe " 7 ' dx 1 c^ x 7 ;" uoe x 7 du oe " dx 7 # # 27. ' 2 dx 1 c 4x ;" u oe 2x du oe 2 dx # # 26. "c ' sin x dx 1cx ; u oe sinc" x du oe dx -- 1cx # "c # 25. ' ax b1b adx tan 2b c 21. 2x 1 ln 2 c bC 22. " 2 S5 5 < b C ln x u oe ln v " du oe v dv ' du oe ln kuk b C oe ln kln vk b C u du oe v dv ' du oe ln kuk b C oe ln k2 b ln vk b C u ' du oe ln kuk b C oe ln k2 b tanc" xk b C u xb ; " u oe 2 b tanc" x du oe x dx 1 b ' u du oe " u# b C oe " asinc" xb# b C # # oe sinc" u b C oe sinc" (2x) b C ' du 1cu ' du 1cu oe sinc" u b C oe sinc" ^ x b C 7 ' 2x 1 dx oe ' 5x2 dx oe ) ) ) ) 20. ' e sec# ae b d); " u oe e ' sec# u du oe tan u b C oe tan ae b b C du oe e d) ) 2 ) ) ) ) ) 19. ' e sin ae b cos# ae b d); " u oe cos ^e) du oe c sin ^e) e d) #" #" 14. ) ' (1 cossin d) b )) ;" u oe 1 b sin ) du oe cos ) d) ' udu oe 2u"# b C oe 21 b sin ) b C du u # # 13. ;" u oe 1 c cos 2) du oe # sin 2) d) " # " ' du oe c 2u b C oe c #(1 c " 2)) b C u cos " " oe c 4 ln kuk b C oe c 4 ln k3 b 4 cos tk b C ' cu# du oe c " u$ b C oe c " cos$ ae b b C 3 3 Chapter 8 Practice Exercises 29. 557 1 c S 3t < 4 1 c S 2t < 3 3 41. 42. x ' sin# x dx oe ' 1 c cos 2x dx oe # c sin42x b C # ' cos# 3x dx oe ' " b cos 6x dx oe x b sin 6x b C # # 12 $ 43. ) ) ) ' sin$ # d) oe ' ^1 c cos# # ^sin # d); oe 44. 2 3 cos$ ) # c 2 cos ) # bC u oe cos ) c ' a1 c u# b u# du oe ' au% c u# b du du oe c sin ) d) ' sin$ ) cos# ) d) oe ' a1 c cos# )b (sin )) acos# )b d); " $ & $ & oe u 5 c u 3 bCoe cos ) 5 c cos ) 3 bC # # 40. ' dv (v b 1)v b 2v oe' d(v b 1) (v b 1)(v b 1) c 1 # # 39. ' dx (x c 1)x c 2x oe' # # 38. b ' t b dt b 5 oe ' (t d(t2) 2) 1 oe tanc" (t b 2) b C 4t b b d(xc1) (x c 1)(x c 1) c 1 # # 37. c " ' y cdy b 8 oe ' (yd(y2) 2) 4 oe # tanc" ^ y c 2 b C c b # 4y # # 36. ' dx 4xcx c3 oe' d(xc2) 1c(xc2) # # 35. ' dx 4x c x oe' d(x c 2) 4 c (x c 2) # # 34. ' 6 dx x4x c 9 oe 3' dx x x c 9 4 # # 33. ' 4 dx 5x25x c 16 oe 4 25 ' dx xx c 16 25 oe " 5 secc" 5x b C 4 oe 2 secc" 2x b C 3 oe sinc" ^ x c 2 b C # oe sinc" (x c 2) b C # # 32. ' 1 bdt 25t ;" u oe 5t du oe 5 dt " 5 du ' 1bu oe " 5 tanc" u b C oe oe secc" kx c 1k b C oe secc" kv b 1k b C ) u oe cos # c2 ' a1 c u# b du oe ) du oe c " sin # d) -- # # # # 31. dt ' 9bt oe " 9 ' dt 1b^t ; uoe du oe " 3 " 3 t dt -- " 3 du ' 1bu oe " 3 tanc" u b C oe # # # 30. ' dt 9 c 4t oe " 3 ' dt ; uoe du oe 2 3 2 3 t dt -- " 2 ' du 1cu # # # ' dt 16 c 9t oe " 4 ' dt ; uoe du oe 3 4 3 4 t dt -- " 3 ' du 1cu oe " 3 sinc" u b C oe " 3 sinc" ^ 3t b C 4 oe " 2 sinc" u b C oe " 2 sinc" ^ 2t b C 3 " 3 t tanc" ^ 3 b C " 5 tanc" (5t) b C 2u 3 c 2u b C 558 45. Chapter 8 Techniques of Integration (tan 2t) asec# 2t c 1b dt oe ' tan 2t sec# 2t dt c ' tan 2t dt; " ln ksec 2tk b C u oe tan t 6 ' au# b 1b du oe 2u$ b 6u b C du oe sec# t dt u oe 2t du oe 2 dt ' tan$ 2t dt oe ' oe " 4 " # tan# 2t c " " " " ' tan u sec# u du c " ' tan u du oe 4 tan# u b # ln kcos uk b C oe 4 tan# 2t b # ln kcos 2tk b C # " # 46. ' 6 sec% t dt oe 6' atan# t b 1b asec# tb dt; " oe 2 tan$ t b 6 tan t b C dx ' 2 sin dxcos x oe ' sin 2x oe ' x dx 2 ' cos 2 csin x oe ' cosdx ; " x 2x # # 47. csc 2x dx oe c " ln kcsc 2x b cot 2xk b C # 48. u oe 2x du oe 2 dx ' du cos u oe ' sec u du oe ln ksec u b tan uk b C oe ln ksec 2x b tan 2xk b C 49. 1 1 1 1 50. ' 3 4 4 cot# t b 1 dt oe ' 3 4 4 csc t dt oe cc ln kcsc t b cot tkd 3 44 oe c ln csc 341 b cot 341 b ln csc 1 b cot 1 4 4 oe c ln 2 c 1 b ln 2 b 1 oe ln 2 b " oe ln 2c1 S2 b "< S2 b 1< #c1 1 1 1 1 1 1 oe c ^c " c " b < " c ^c " ` oe 2 # # # # 52. 1 1 1 1 oe (2 c 0) c (0 c 2) oe 4 53. 1 1 1 54. 57. 58. 59. b ' 4x c 13 dx oe ' <(2x b 1) b #x 4c 1 ` dx oe x b x# b 2 ln k2x c 1k b C 2x 8 ' 2xcdx oe ' ^2 b x c 4 dx oe 2x b 8 ln kx c 4k b C x 4 c1 ' 2y b 4 dy oe ' y # # # 2y dy y b4 # c' dy y b4 oe ln ay# b 4b c # # # 56. " # y tanc" ^ # b C # # ' 9xbdx x $ 9 oe ' ' x ax xbbb9c 9x " dx oe ' ^x c # # 55. dx dx x ' xx b 4 oe x c ' x4 b 4 oe x c 2 tanc" ^ # b C # 1$ 1 oe c2 csin td 2 b 2 csin td #1# oe c2 (c1 c 0) b 2 [0 c (c1)] oe 22 $1 9x x b9 dx oe x # c 9 # ln a9 b x# b b C 1 1 1 ' 2 1 b cos 2t dt oe 2 ' 2 kcos tk dt oe c2 ' 3 1 1 1 1 1 1c 2 2 2 cos t dt b 2 '3 2 cos t dt 2 1 1c ' 2 1 c cos 2t dt oe 2 ' 2 ksin tk dt oe 22 '0 sin t dt oe 'c22 cos t" 2 2 0 oe 22 [0 c (c1)] oe 22 1# 1 '02 1 c sin# x dx oe '02 cos x dx oe '0 cos x dx c ' 2 cos x dx oe <2 sin x ` 0 c <2 sin x ` # # # # # # 1 1 1 51. 2 1 1 1 sin 2x dx c 1 1 '0 1 c cos# 2x dx oe '0 ksin 2xk dx oe '0 1 1 1 1 ' 2 4 csc# y c 1 dy oe ' 2 4 cot y dy oe cln ksin ykd 2 4 oe ln 1 c ln " 2 oe ln 2 oe ln S3 b 2 2< 2 ' sin 2x dx oe c < cos 2x ` 0 b < cos 2x ` # # 2 2 Chapter 8 Practice Exercises # # # 559 60. 61. dy " ' yy b 41 dy oe ' yy b 1 b 4 ' y dy 1 oe # ln ay# b 1b b 4 tanc" y b C b b 65. ' sec (5 c 3x) dx; " y oe 5 c 3x dy oe c3 dx " ' sec y Sc dy < oe c " ' sec y dy oe c 3 ln ksec y b tan yk b C 3 3 oe c " ln ksec (5 c 3x) b tan (5 c 3x)k b C 3 66. 67. 68. ' x csc ax# b 3b dx oe " ' csc ax# b 3b d ax# b 3b oe c " ln kcsc ax# b 3b b cot ax# b 3bk b C # # ' cot ^ x dx oe 4 ' cot ^ x d ^ x oe 4 ln sin ^ x b C 4 4 4 4 ' tan (2x c 7) dx oe " ' tan (2x c 7) d(2x c 7) oe c " ln kcos (2x c 7)k b C oe " ln ksec (2x c 7)k b C # # # ' x1 c x dx; " u oe 1 c x du oe c dx oe 2 5 69. c ' (1 c u)u du oe ' ^u$# c u"# du oe c & $ (1 c x)&# c 2 (1 c x)$# b C oe c2 3 S1 c x< 3 S1 c x< 5 70. ' 3x2x b 1 dx; " u oe 2x b 1 du oe 2 dx oe 3 10 ' 3 ^ u c 1 u " du oe 3 ' ^u$# c u"# du oe 3 2 u&# c 3 2 u$# b C # # 4 4 5 4 3 c $ & (2x b 1)&# c " (2x b 1)$# b C oe # z oe tan ) dz oe sec# ) d) b " # 3c2 3c1 3 ^2x b 1 10 ^2x b 1 # 71. ' z# b 1 dz; " oe oe sec ) tan ) 3c1 sin ) 2 cos ) # ' tan# ) b 1 sec# ) d) oe ' sec$ ) d) (FORMULA 92) z z b 1 # # ' sec ) d) b ln ksec ) b tan )k b C oe z oe 4 tan ) dz oe 4 sec# ) d) b " # #" # oe z 16 a16 b z b bC # $ 72. # ' a16 b z# bc$# dz; " 4 " " ' 64sec ))dd)) oe 16 ' cos ) d) oe 16 sin ) b C oe sec # # 64. x dx cos dx ' cotcotb csc x oe ' cos xxb 1 oe ' (cos1x)(1 c cos x) dx oe ' cos x c 1 b sin x dx x c cos x sin x x) " dx ' d(sin x c ' sin x b ' dx oe c sin x b cot x b x b C oe x b cot x c csc x b C oe sin # # # # # 63. c x dx sin dx ' tantanb sec x oe ' sin xxb 1 oe ' (sin1x)(1sin sin x) dx oe ' sin x c 1 bxcos x dx x c x cos x) " dx ' d(cos x c ' cos x b ' dx oe cos x c tan x b x b C oe x c tan x b sec x b C oec cos # # # # # 62. t 1ct # b ' 2t 1 c t # dt oe ' 2t dt 1 c t b' dt t oe c21 c t# b ln ktk b C # # # ' t b 2 4ct dt oe ' t dt 4 c t b 2' dt 4 c t t oe c4 c t# b 2 sinc" ^ # b C 2 5 u&# c 2 u$# b C 3 --bC bC ln z b 1 b z# b C z 1616 b z bC 560 73. Chapter 8 Techniques of Integration y 1 b^ 5 1 b tan ) # oe ln y b 25 b y# b C 74. 1 b S 3y 5 < 75. " 3 ln 25 b 9y# b 3y b C ;" # sin x # 78. ' 4 c x# dx; " x oe 2 sin ) dx oe 2 cos ) d) ' 2 cos ) 2 cos ) d) oe 2 ' (1 b cos 2)) d) oe 2 ^) b " sin 2) b C # # oe 2) b 2 sin ) cos ) b C oe 2 sinc" ^ x b x1 c ^ x b C oe 2 sinc" ^ x b # # # 79. 12 oe c 12 b C oe c sin ) b C oe c 12 x b C u x c1 # 81. ' ww c 1 dw; " # w oe sec ) tan ' ^ sec ) sec ) tan ) d) oe ' tan# ) d) oe ' asec# ) c 1b d) ) dw oe sec ) tan ) d) oe tan ) c ) b C oe w# c 1 c secc" w b C z oe 4 sec ) dz oe 4 sec ) tan ) d) sec ' 4 tan )44sec )) tan ) d) oe 4 ' 82. ' z zc 16 dz; " # tan# ) d) oe 4(tan ) c )) b C z oe z# c 16 c 4 secc" ^ 4 b C 83. u oe ln (x b 1), du oe x ' ln (x b 1) dx oe x ln (x b 1) c ' x b 1 dx oe x ln (x b 1) c ' dx b ' x dx 1 oe x ln (x b 1) c x b ln (x b 1) b C" b dx xb1 ; dv oe dx, v oe x; oe (x b 1) ln (x b 1) c x b C" oe (x b 1) ln (x b 1) c (x b 1) b C, where C oe C" b 1 # # $ #$ # 80. ' 12 dx ax c 1 b ;" x oe sec ) dx oe sec ) tan ) d) ) cos ' 12 sectan tan ) d) oe ' 12 sin ) d) ; " ) ) # oe ln ksec ) b tan )k b C" oe ln x b ^ x c 1 b C" oe ln x b 3 3 # # ' dx # x c9 ;" x oe 3 sec ) ' dx oe 3 sec ) tan ) d) # "c oe " # )c # 77. ;" " # sin ) cos ) oe c x 1 c x # # # ' x dx 1cx x oe sin ) dx oe cos ) d) # # Note: Ans cx 1 c x 3 c 2 1 c x# b C by another method 3 cos " ' sin )cos ) ) d) oe ' sin# ) d) oe ' 1 c cos 2) d) oe " ) c 4 sin 2) b C # # bC 3 sec ) tan ) d) 9 sec ) c 9 $ cu oe cos )d c ' a1 c u# b du oe cu b # 76. ;" $ $ ' x dx 1cx x oe sin ) dx oe cos ) d) cos ' sin )cos ) ) d) oe ' sin$ ) d) oe ' a1 c cos# )b (sin )) d); u 3 # # # ' dx x 1 c x x oe sin ) dx oe cos ) d) cos d ' sin ))cos) ) oe ' csc# ) d) oe c cot ) b C oe c1x c x # # # ' 25dy 9y b oe " 5 ' dy oe " 3 ' du 1bu oe " 3 ln 1 b u# b u b C" from Exercise 73 bC b C oe c cos ) b " 3 " cos$ ) oe c1 c x# b 3 a1 c x# b # x 4 c x 2 bC oe' 3 sec ) tan ) d) 3 tan ) oe ' sec ) d) b C" oe ln x b x# c 9 b C u oe sin ) du oe cos ) d) x c 9 3 ' 12u du # oe ln ksec ) b tan )k b C" oe ln 1 b u# b u b C" oe ln 1 b ^ y b y b C" oe ln 5 5 # # # # # dy ' 25 b y oe " 5 ' dy oe' du 1 b u u oe tan ) , <u oe y ` ; " 5 du oe sec# ) d) ' sec ) d) oe ' sec ) d) 25 b y b y 5 b C" $# bC Chapter 8 Practice Exercises 84. u oe ln x, du oe " " " ' x# ln x dx oe 3 x$ ln x c ' 3 x$ ^ x dx oe x3 3 dx 1 b 9x dx x 561 ; dv oe x# dx, v oe " 3 x$ ; $ ln x c x 9 bC oe x tanc" (3x) c " 6 ln a1 b 9x# b b C c dx 4 c x oe x cosc" ^ x c 4 c x# b C oe x cosc" ^ x c 21 c ^ x b C # # # 87. ex b (x b 1)# ex c 2(x b 1) ex b 2 ex 0 88. b x# c 2x b 2 0 sin (1 c x) cos (1 c x) c sin (1 c x) c cos (1 c x) ' (x b 1)# ex dx oe c(x b 1)# c 2(x b 1) b 2d ex b C ' x# sin (1 c x) dx oe x# cos (1 c x) b 2x sin (1 c x) c 2 cos (1 c x) b C 89. u oe cos 2x, du oe c2 sin 2x dx; dv oe ex dx, v oe ex ; I oe ' ex cos 2x dx oe ex cos 2x b 2 ' ex sin 2x dx; u oe sin 2x, du oe 2 cos 2x dx; dv oe ex dx, v oe ex ; I oe ex cos 2x b 2 'ex sin 2x c 2 ' ex cos 2x dx" oe ex cos 2x b 2ex sin 2x c 4I I oe c c 90. u oe sin 3x, du oe 3 cos 3x dx; dv oe e c c 2x # 92. x dx 3 3 " " ' x b 4x b 3 oe # ' xdx3 c # ' x dx 1 oe # ln kx b 3k c # ln kx b 1k b C b b # 91. x dx dx ' x c 3x b 2 oe ' x2c 2 c ' x dx 1 oe 2 ln kx c 2k c ln kx c 1k b C c c c c c Ioe 4 13 ^c " e # 2x sin 3x c 3 e 4 2x 2 cos 3x b C oe c 13 e 2x sin 3x c 3 13 e 2x cos 3x b C c c c c c Ioec"e # 2x sin 3x b 3 'c " e # # 2x cos 3x c 3 # 'e 2x sin 3x dx" oe c " e # c c u oe cos 3x, du oe c3 sin 3x dx; dv oe e 2x dx, v oe c " e # c Ioe'e dx, v oe c " e # 3 2 2x sin 3x dx oe c e " # 2x sin 3x b # ' cosc" ^ x dx oe x cosc" ^ x b ' x dx # # # 86. u oe cosc" ^ x , du oe # ; dv oe dx, v oe x; 4 cx # dx ' tanc" 3x dx oe x tanc" 3x c ' 13x 9x b # 85. u oe tanc" 3x, du oe ; dv oe dx, v oe x; ;" y oe 1 b 9x# x tanc" 3x c dy oe 18x dx " 6 ;" 'e $ ' dy y y oe 4 c x# x cosc" ^ x c # dy oe c2x dx # " # dy ' y ex cos 2x 5 b 2ex sin 2x 5 bC 2x ; 2x cos 3x dx; 2x ; 2x sin 3x c 3 e 4 2x cos 3x c 9 I 4 562 93. Chapter 8 Techniques of Integration # oe 96. 97. 98. 99. " 3 b c ln cos ) c 2 b C oe c " ln cos ) b 1 b C cos ) 1 3 cos ) 2 dx x ' x4xbdx oe ' x4 b 4 oe 2 tanc" ^ # b C 4x 3) dv " ' (v bc 8v oe # ' 2v ' & # $ 3 Sc 4v b 5 8(v c 2) oe " 16 ln (v c 2)v (v b 2) b C dv vc2 oe b " 3 ln kx c 2k b C oe " 3 ln x b 1 c " b C xb1b1 # 107. ' dx x ^3 x b 1 u oe x b 1 ; du oe 2dxb 1 x dx oe 2u du 2 3 " " " " ' au ucdub u oe 3 ' u du 1 c 3 ' u du 1 oe 3 ln ku c 1k c 3 ln ku b 1k b C 1 c b # # 106. # $ ' 2x b x c 21x b 24 dx oe x b 2x c 8 2 # x c 3x b 3 ln kx b 4k # oe x # c 9 # ln kx b 3k b 3 # ln kx b 1k b C <(2x c 3) b x ` x b 2x c 8 ' # # 105. ' x xbb 4x 3 dx oe ' ^x c x b3x b 3 dx oe ' 4x b 4x # $ $ $ 104. ' x b 1 dx oe ' ^1 b xx b "x dx oe ' '1 b x(x "c 1) " dx oe ' dx b ' x dx 1 c ' dx oe x x cx c c x $ # oe x # b 4 3 ln kx b 2k b 2 3 ln kx c 1k b C b ln kx c 1k c ln kxk b C # # 103. 2 4 ' x xbb x 2 dx oe ' ^x b x b2x c 2 dx oe ' x dx b 3 ' x dx 1 b 3 ' x dx 2 xc x c b # $ # # # % 102. dt dt " " " ' t ct tdtc 2 oe " ' t t c 2 c 3 ' t t b 1 oe 6 ln kt# c 2k c 6 ln at# b 1b b C 3 # # # % 101. dt " " ' t b 4t b 3 oe " ' t dt 1 c # ' t dt 3 oe # tanc" t c # b b " # 3 t tanc" S 3 < b C oe x dx b 3 # 9 ' x dx 1 c # ' xdx3 b b dx oe ' (2x c 3) dx b # 100. dv ' (v c (3v cc7) dv c 3) oe ' (c2) 1 b ' 1)(v 2)(v vc # 4x ' 3x xb b xb 4 dx oe ' $ # 4 x dx c ' xc4 x b1 b # cos ) d ' sin ) b sin )) c 6 ; csin ) oe xd # ' dx x bxc6 dx oe 4 ln kxk c " 8(v b #) < # # 95. ' cos )sin ) d)) c 2 ; ccos ) oe yd b cos c' # $ # 94. b1 2 ' x x(xc1) dx oe ' ^ xc1 c 2 c x" dx oe 2 ln xc" b " b C oe c2 ln kxk b " b 2 ln kx c 1k b C x x x x # ' x(xdx 1) b oe ' S" c x 1 xb1 b c1 (x b 1) < dx oe ln kxk c ln kx b 1k b " xb1 bC dy y byc2 oec"' 3 dy yc1 b " 3 yb2 " ' y dy# oe 3 ln y c 1 b C b oe " 5 " " c ' xdx 2 c 5 ' xdx 3 oe 5 ln sin ) b 2 b C sin ) 3 c b " # ln ax# b 1b b 4 tanc" x b C dv oe c 3 ln kvk b 8 5 16 ln kv c 2k b " 16 ln kv b 2k b C b' dv vc3 oe ln (v c 2)(v c 3) b C (v c 1) " # tanc" t c 3 6 tanc" t 3 bC " 3 2 ' x dx# b 3 ' x dx 4 c b Chapter 8 Practice Exercises 3u du u (1 b u) # $ 563 s ds oe du ub1 s oe ln es b 1 c " es b 1 b 1 u c1 # bC d a16 c y b 16 c y # # oe c " ln k4 c x# k b C # 114. (a) (b) # 4t c1 # oe 117. " 9 ln kxk c oe " 6 119. 120. ' sin3 x cos4 x dx oe ' cos4 xa1 c cos2 xbsin x dx oe ' cos4 x sin x dx c ' cos6 x sin x dx oe c cos x b cos x b C 5 7 5 7 ' cos5 x sin5 x dx oe ' sin5 x cos4 x cos x dx oe ' sin5 x a1 c sin2 xb2 cos x dx oe ' sin5 x cos x dx c 2' sin7 x cos x dx b ' sin9 x cos x dx oe sin x c 2sin x b sin x b C 6 8 10 6 8 10 # 118. ' dx 9cx # dx ' 9cx # 116. " " " " " " ' x a9dx x b oe 9 ' dx b 18 ' 3 dx x c 18 ' 3 dx x oe 9 ln kxk c 18 ln k3 c xk c 18 ln k3 b xk b C c x c b " 18 ln k9 c x# k b C " " " " xb3 ' 3 dx x b 6 ' 3dxx oe c 6 ln k3 c xk b 6 ln k3 b xk b C oe 6 ln x c 3 b C c b ;" x oe 3 sin ) dx oe 3 cos ) d) x ' 3 cos ) d) oe ' d) oe ) b C oe sinc" 3 b C 3 cos ) # # 115. ' 9xcdx x ;" u oe 9 c x# " c#' du oe c2x dx du u " oe c # ln kuk b C oe ln " u b C oe ln " 9 c x bC # ; <t oe sec )` # " # " ' t dt " # # ' t dt # 4t c 1 oe " 8 a ' d4t c 1b 4t c 1 oe " 4 4t# c 1 b C sec ) tan ) sec ) d) tan ) ' # # (b) oe " 4 ' sec# ) d) oe tan ) b C oe 4t4 c 1 b C 4 # ; cx oe 2 sin )d # # 113. (a) # ' 4xcdx x ' 4xcdx x " oec#' # (b) x dx 4 b x d a4 c x b 4cx " oe c # ln k4 c x# k b C 2 ' 2 sin4)coscos ) d) oe ' tan ) d) oe c ln kcos )k b C oe c ln S 4 2c x < b C ) # ' ; cx oe 2 tan yd # # 112. (a) 4bx 4bx # ' x dx # (b) sin x cos x dx cos x oe " # a ' d4 b x b oe 4 b x# b C ' 2 tan y2 sec y dy oe 2 ' sec y tan y dy oe 2 sec y b C oe 4 b x# b C 2 sec y # y ' 16dy y c # 111. (a) y ' 16dy y c oec"' # oe c16 c y# b C oe c4 cos x b C oe c 416 c y 4 ; cy oe 4 sin xd 4 ' # 110. ' ds es b 1 s u oe esb 1 e du oe ds ; 2 e s b1 ds oe 2u du du uc ' u a2u du1b oe 2 ' (u b 1)(u c 1) oe ' u du 1 c ' u du 1 oe ln u b " b C 1 c b u c b C oe c16 c y# b C c 109. ' e ds 1 ; du oe es ds c u oe es c 1 u ' u(udu 1) oe c' udu1 b ' du oe ln u b 1 b C oe ln e ec " b C oe ln k1 c e u b b s $ $# $ 108. du u(1 b u) u oe 3 ln u b 1 b C oe 3 ln 1bx b C $ dx ' x ^1 b x ; u oe x dx ' du oe 3x # dx oe 3u du $ oe 3' x s kbC 564 121. 122. Chapter 8 Techniques of Integration ' tan4 x sec2 x dx oe tan x b C 5 5 ' tan3 x sec3 x dx oe ' asec2 x c 1b sec2 x sec x tan x dx oe ' sec4 x sec x tan x dx c ' sec2 x sec x tan x dx oe sec5 x 5 c sec3 x 3 bC 123. " ' sin 5) cos 6) d) oe " ' asinac)b b sina11)bb d) oe " ' sinac)b d) b " ' sina11)b d) oe " cosac)b c ## cos 11) b C # # # # oe " cos ) c # " ## cos 11) b C d) b t 4 " # 1 ' cos 6) d) oe " ) b 12 sin 6) b C # 124. 125. ' cos 3) cos 3) d) oe " ' acos 0 b cos 6)b d) oe " ' # # ' 1 b cos^ 2t dt oe ' 2 cos t 4 dt oe 42 sin bC 126. ' et tan2 et b 1 dt oe ' k sec et k et dt oe lnk sec et b tan et k b C (x) oe 24xc& which is decreasing on [1 3] maximum of f % (x) on [1 3] is f % (1) oe 24 M oe 24. Then % c " " kEs k Y 0.0001 ^ 3180" ^ 2 (24) Y 0.0001 ^ 768 ^ n Y 0.0001 n Y (0.0001) ^ 180 n% 10,000 ^ 768 n 180 768 180 f n 14.37 n 16 (n must be even) % % 127. kEs k Y 3c" 180 % (~x)% M where ~x oe 3c" n oe 2 n ; f(x) oe " x oe xc" f w (x) oe cxc# f ww (x) oe 2xc$ f'''(x) oe c6xc% 128. kET k Y # 1c0 12 (~x)# M where ~x oe # 2 3n Y 10c$ oe 1c0 6 3n # 1000 n ~x # 1c0 n # oe " n ;0 2000 3 Y f ww (x) Y 8 M oe 8. Then kET k Y 10c$ n 25.82 n 26 xi 0 1/6 1/3 1/2 21/3 51/6 1 xi 0 1/6 1/3 1/2 21/3 51/6 1 f(xi ) 0 1/2 3/2 2 3/2 1/2 0 f(xi ) 0 1/2 3/2 2 3/2 1/2 0 m 1 2 2 2 2 2 1 m 1 4 2 4 2 4 1 " 12 ^ " # (8) Y 10c$ n 129. ~x oe bca n 6 oe 1 6 oe 1 1# ; x! x" x# x$ x% x& x' x! x" x# x$ x% x& x' ! mf(xi ) oe 12 T oe ^ 1 (12) oe 1 ; 12 i 0 mf(xi ) 0 1 3 4 3 1 0 mf(xi ) 0 2 3 8 3 2 0 & ! mf(xi ) oe 18 and i 0 1 S oe ^ 18 (18) oe 1 . n 6.38 n 8 (n must be even) 131. yav oe oe oe " 365 c 0 " <^ 21 21 c37 ^ 365 cos < 365 (365 c 101)` b 25(365) c ^c37 ^ 365 cos < 365 (0 c 101)` b 25(0)` 365 21 21 37 21 37 21 37 21 21 c 21 cos ^ 365 (264) b 25 b 21 cos ^ 365 (c101) oe c 21 ^cos ^ 365 (264) c cos ^ 365 (c101) b 21 " 21 '0365 <37 sin ^ 365 (x c 101) b 25` dx oe 365 <c37 ^ 365 cos ^ 365 (x c 101) b 25x` $'& 21 ! % 130. f % (x) Y 3 M oe 3; ~x oe oe 6 ~x 3 oe 1 18 oe 2c" n oe " n % c . Hence kEs k Y 10c& ^ 2180" ^ " (3) Y 10c& n " 60n Y 10c& n% 10 60 25 Chapter 8 Practice Exercises 37 c #1 (0.16705 c 0.16705) b 25 oe 25 F " 675c20 675 " " '20 c8.27 b 10c& a26T c 1.87T# bd dT oe 655 <8.27T b 103 565 " 655 [(5582.25 b 59.23125 c 1917.03194) c (165.4 b 0.052 c 0.04987)] 5.434; 26b676 b 4(1.87)(283,600) #(1.87) 8.27 b 10c& a26T c 1.87T# b oe 5.434 1.87T# c 26T c 283,600 oe 0 T 396.45 C 133. (a) Each interval is 5 min oe 1 2.5 24 c 1 12 hour. 29 12 b 2a2.4b b 2a2.3b b b 2a2.4b b 2.3 d oe (b) a60 mphb^ 12 hours/gal 24.83 mi/gal 29 134. Using the Simpson's rule, ~x oe 15 ~x oe 5; 3 ! mf(xi ) oe 1211.8 Area a1211.8ba5b oe 6059 ft2 ; The cost is Area a$2.10/ft2 b a6059 ft2 ba$2.10/ft2 b oe $12,723.90 the job cannot be done for $11,000. 2.42 gal x! x" x# x$ x% x5 x' x( x) xi 0 15 30 45 60 75 90 105 120 f(xi ) 0 36 54 51 49.5 54 64.4 67.5 42 1 # b oe c1 b 0 oe c 1 137. 2 " oe lim <ln ^ 4b b 1 c b ` c (ln 1 c 1 c ln 3) oe ln c b_ b_ b_ # # # 143. dx 4x b 9 dx 4x b 9 oe " # _ _ _c _ ' _c 142. '0 xe3x dx oe b c_ lim " " < x e3x c 9 e3x ` 0 oe c 9 c 3 b b c_ lim " " " ^ b e3b c 9 e3b oe c 9 c 0 oe c 9 3 oe 2 '0 '0 dx x b9 4 oe " # b lim _ 2 < 3 tanc" ^ 2x ` b oe 3 0 " # b lim _ 2 < 3 tanc" ^ 2b ` c 3 c c c c c c c 141. _ '0 x# e x dx oe lim ccx# e x c 2xe x c 2e x d 0 oe lim acb# e b c 2be b c 2e b b c (c2) oe 0 b 2 oe 2 b # # $ 140. 3v c 1 4v c v _ _ '1 # 139. 2 du u c 2u du uc2 dv oe '1 ^ " b v " v _ _ _ '3 oe '3 c '3 du u b oe lim <ln u c 2 ` 3 oe lim <ln b c 2 ` c ln 3 c 2 oe 0 c ln ^ " oe ln 3 u b 3 3 b_ c 4 4v c 1 dv oe lim <ln v c b_ " 4 &$ c &$ &$ lim ) _ ()b1) &$ ) d) ) diverges _ _ oe 1 and '2 1 ' d) () b 1) &$ &$ c &$ c &$ c 138. d) 2 () b 1) d) 2 () b 1) d) () b 1) _ c _ ' oe' 1 b' 2 b '2 d) () b 1) converges if each integral converges, but diverges b_ " v c ln (4v c 1)` 1 3 4 b b 1 b ln 3 oe 1 b ln b b 1 $# $# $# c $# ' 11 ydy c oe' 0 dy y b '0 1 dy y oe 2 '0 1 dy y oe 2 3 lim b! <y"$ ` 1 oe 6 S1 c lim b b! b"$ < oe 6 " 3 tanc" (0) # " b " b b b 136. b! lim cx ln x c xd 1 oe (1 ln 1 c 1) c lim b b! cb ln b c bd oe c1 c lim b! ln b Sb< oe c1 c lim b! Sc " '01 ln x dx oe c c # c # 135. '03 dx 9cx oe lim b3 '0b dx 9cx oe lim b3 <sinc" ^ x ` b oe lim 3 0 b3 b 0 sinc" ^ 3 c sinc" ^ 3 oe & & 132. av(Cv ) oe T# c 0.62333 10 T$ ` #! '(& m 1 4 2 4 2 4 2 4 1 c0oe 1 # mf(xi ) 0 144 108 204 99 216 128.8 270 42 Sb< < 566 oe 144. Chapter 8 Techniques of Integration " # ^2 1 c 0 oe 3 # 4 dx x b 16 # 1 6 b oe 2 lim <tanc" ^ x ` 0 oe 2 S lim <tanc" ^ b ` c tanc" (0)< oe 2 ^ 1 c 0 oe 1 4 4 # I oe 1 b 0 c I 2I oe 1 I oe 147. ln z z e b_ " # b_ converges # # # ln z z ln z z oe _ diverges c c c 1 151. 154. ' cos x dx; x ' dx c2x c x u oe x du oe #dxx -- d(x b 1) 1 c (x b 1) ' cos uu2u du oe 2' cos u du oe 2 sin u b C oe 2 sin x b C # 157. 1bu # ' du # 156. ' (t c 1) dt ' " t c 2t u oe t# c 2t du oe (2t c 2) dt oe 2(t c 1) dt ; cu oe tan )d # # 155. oe' oe sinc" (x b 1) b C " # du ' u oe u b C oe t# c 2t b C ' sec ))d) oe ln ksec ) b tan )k b C oe ln 1 b u# b u b C sec # # oe ln ksin )k c " # sin# ) b C oe ln xx b 1 c " S xx b 1 < b C # % # # 153. ;" # # $ # ' dx x ax b 1 b x oe tan ) dx oe sec# ) d) # # 152. sec ) d ' tan ) sec )) oe ' cos ))d) oe ' S 1 c sin) ) < d(sin )) sin sin # b2 ' x cbx2 dx oe c ' ^x b 4x c 4 dx oe c ' 4 x $ #$ oe 2x 3 c x b 2x c 2 ln ^1 b x b C x dx c 3 # 5 ' x dx 2 c # ' x dx 2 oe c x# b c # x ' 1 bdx x ; # _c _ ' dx x a1 b e x b u oe x du oe #dxx -- b 1 ex diverges 2 2 ' u 12u udu oe ' ^2u# c 2u b 2 c 1 b u du oe 3 u$ c u# b 2u c 2 ln k1 b uk b C b c 3 # ln kx b 2k c # # # b a# " x 0 'x lim " x0 # Sx < # " oe lim x a1 b e x b x oe lim a1 b ex b oe 2 and '0 x0 1 dx x diverges # # # c # _c # _c 150. dx x a1 b e x b dx x a1 b e x b dx x a1 b e x b dx x a1 b e x b _ c _ ' oe' 1 b' 0 b '0 1 b '1 c _c c c _c 149. 2 dx ex b e x 2 dx ex b e x 4 dx ex converges _ _ _ _ ' oe 2'0 '0 ' 2 dx ex b e c 148. 0 e t t _ Y e t for t 1 and '1 e t dt converges _ # _ _ '1 dz oe '1 dz b 'e dz oe ' (ln2z) " b lim ' (ln#z) " oe S 1 c 0< b lim ' (ln2b) c " " 2 # 1 e b b_ e b_ '1 e t t dt converges x converges dx x a1 b e x b ; '01 x a1dx e b diverges b x 5 # ln kx c 2k b C c c c c u cos u du oe lim cce u u sin u du oe 1 b lim ce u _ _ c 146. I oe '0 e cos ud b c '0 e 0 # # 145. lim ) _ ) b 1 ) d) ) _ _ oe 1 and '6 # _ _ _c _ ' oe 2'0 4 dx x b 16 b_ b_ diverges '6 d) ) b 1 diverges sin ud b c '0 ae u b cos u du 0 Chapter 8 Practice Exercises 158. 159. 567 ' et cos et dt oe sin et b C # ' 2 c cos x b sin x dx oe ' sin x # # # 2 csc# x dx c ' cos x dx sin x b ' csc x dx oe c2 cot x b " sin x c ln kcsc x b cot xk b C oe c2 cot x b csc x c ln kcsc x b cot xk b C 160. 161. 162. 163. b ) c 1 0 164. sin ' cos )) d) oe ' 1 c cos) ) d) oe ' sec# ) d) c ' d) oe tan ) c ) b C cos dv ' 819c v " # " " bv " ' v dv 9 b 12 ' 3 dv v b 1"# ' 3 dv v oe 12 ln 3 c v b 6 tanc" v b C b c b 3 3 # # # oe x ' 1cossindxx oe ' 1d(sin x)x oe tanc" (sin x) b C b b sin cos (2) b 1) " # sin (2) b 1) c " cos (2) b 1) 4 ) ' ) cos (2) b 1) d) oe # sin (2) b 1) b " cos (2) b 1) b C 4 166. ' 1 b ) oe 4 3 S1 b ) < $# c 4 S1 b ) < "# bCoe4 OE1 b )9 3 167. dx ' 2 sinsec xx ; x y oe x dy oe 2dxx -- y ' 2 sin sec2y dy oe ' 2 sin 2y dy oe c cos (2y) b C oe c cos ^2x b C y y 169. 170. ' dy sin y cos y oe' 2 dy sin 2y oe ' 2 csc (2y) dy oe c ln kcsc (2y) b cot (2y)k b C c ' ) cd2)) b 4 oe ' () c d) b 3 oe 3 tanc" S ) " < b C 1) 3 tan ' cos xx dx oe ' tan x sec# x dx oe ' tan x d(tan x) oe " tan# x b C # # # 3 # # 172. ' dr (r b 1)r b 2r # 171. oe' d(r b 1) (r b 1)(r b 1) c 1 oe secc" kr b 1k b C # # # % % 168. # # # ' xx cdx oe ' ^x b x 16x16 dx oe x# 16 c & b ' ^ x 2x 4 c c 2x x b4 dx oe $ # oe x # b 2x b 3 ln kx c 1k c ; " xc1 bC d) 2 ) d) oe 2(x c 1) dx x oe 1 b ) d) dx oe dx 4 ' 2(x c 1) dx oe 2 ' x dx c 2 ' x oe 3 x$# c 4x"# b C x c 1 b ) b C x # x b ln x c4 b4 bC # # # 165. x dx c ' x c 2x b 1 oe ' ^x b 2 b x 3x2x 2 1 dx oe ' c b $ # _ '2 dx (x c 1) b " " oe lim < 1 c x ` 2 oe lim < 1 c b c (c1)` oe 0 b 1 oe 1 b_ # # % b_ (x b 2) dx b 3 ' dx xc1 b' dx (x c 1) 568 173. Chapter 8 Techniques of Integration (r ' b 2) dr cr c 4r # % 177. 178. ' (15)2xb1 dx oe " ' (15)2xb1 d(2x b 1) oe " S 15 15 < b C # # ln 2x 1 179. ' x dx oe 2 2cx ;" 3 yoe2cx c' dy oe c dx $ (2 c y) dy y oe 2 3 y$# c 4y"# b C oe S2 c x< c 22 c x-- b C sin cos ' cos )sin )) d) oe ' a1 csin ))b d) oe ' csc# ) d) c ' d) oe cot ) c ) b C # # # 182. ' ln x c 1 dx; y oe oe " # dy oe ' 2y ln y dy oe y# ln y c ' y dy oe y# ln y c " y# b C oe (x c 1) ln x c 1 c " (x c 1) b C" # # c(x c 1) ln kx c 1k c xd b ^C" b " oe # " 3 " # 183. 184. ' )# tan a)$ b d) oe x dx 8 c 2x c x # " ' tan a)$ b d a)$ b oe 3 ln ksec )$ k b C d ax b 1 b # 9 c ax b 1 b # 186. 187. 188. 0 1 1 '0 10 1 b cos 5) d) oe 2 ' # # 9 c 4t # ' t dt oec"' 8 d a9 c 4t b 9 c 4t " oe c 4 9 c 4t# b C 10 cos ^ 5#) d) oe 2 2 5 <sin ^ 5#) ` 1"! oe ! # # # # dx oe " # d a x# b oe " # ^x# ex c ex b C oe # # ' x$ ex ' x# ex # # # # 185. 1 ' z azzbb 4b dz oe " ' ^ " b z" 4 z c zb1 z b4 dz oe " 4 ln kzk c # # % ' oe " # ' # # 181. c ' y cdy b 2 oe ' (yd(y1) 1) 1 oe tanc" (y c 1) b C c b 2y # oe csinc" v c # 180. # ' 1vc v dv; cv oe sin )d 1 c v v bC x c 1 dx 2 x c 1 -- ' ln y 2y dy; u oe ln y, du oe cx ln kx c 1k c x c ln kx c 1kd b C oe " # sinc" S x b1 3 < bC " 4z c " 8 ln az# b 4b c bC ax c 1 b e x # 1 b 1 4 4 cos 2x dx oe 'c sin 2x" 1 1 1 1 ' 2 1 b cos 4x dx oe c2 ' # # # # # 176. ' dx ax c 1 b oe' dx a1 c x b oe x 2 a1 c x b # # 175. 2) ' (1sincos d))) b 2 " oec#' # # 174. d(1 b cos 2)) (1 b cos 2)) oe b 2 " #(1 b cos 2)) # # ' 4ybdy y oe " # ' d ay b 4 b ay b oe " 4 # oe' (r b 2) dr 4 c (r b 2) ;" u oe % c (r b 2)# c' du oe c2(r b 2) dr du 2 u oe cu b C oe c4 c (r b 2)# b C tanc" S y < b C # bCoe " 4 sec# ) b C " 4 xb ln x c " b C (FORMULA 19) 1 2 # 2 4 oe 2 # 2 3 (2 c x)$# c 4(2 c x)"# b C dy y ; dv oe 2y dy, v oe y# " 8 tanc" z # bC 2 2 5 ^sin 1 4 c 0 oe 2 5 Chapter 8 Practice Exercises # # # # 569 189. ) ) ' 1cot sind)) oe ' (sin )cos1)bdsin )b ; " b )a x oe sin ) dx oe cos ) d) dx ' x a1dx x b oe ' dx c ' xx b 1 b x oe ln ksin )k c " ln a1 b sin# )b b C # # # 190. u oe tanc" x, du oe # "c 191. ' tan2yy dy ; <y oe x` 'e oe 2t 2x ' tan x2x dx oe ln ksec xk b C oe ln sec y b C 192. et dt dx t (x b 1)(x b 2) b 3et b 2 ; ce oe xd t ln x b " b C oe ln ^ et b " b C xb# e b# ' oe' dx xb1 c' oe c) b ln 194. ) b2 ) c2 bC tan# x dx oe ' asec# x c 1b dx oe tan x c x b C ' 1 c cos 2x dx oe ' 1 b cos 2x # "c asin ' cos xb dx 1cx 195. oe ln kcsc (2x) b cot (2x)k b C 197. 198. ' sin # x # cos x # dx oe ' " # sin ^ x b x dx oe # # x dx ax b 2 b " # ' sin x dx oe c " cos x b C # c" 199. 200. ' et dt 1 b et oe ln a1 b et b b C # # ' tan$ t dt oe ' _ (tan t) asec# t c 1b dt oe b_ cb oe lim ^ 2e2b c b_ # " 4e2b c ^0 c " oe 4 " 4 202. ' 3 b sectanxxb sin x dx oe 3 ' cot x dx b ' sec xxdx b ' tan (sin v) cot v v dv ' ln (sindv oe ' (sincosln (sin v) ; " u oe ln cos v dv v) v) du oe sin v # cos x dx oe 3 ln ksin xk b ln ktan xk b sin x b C 203. ' du oe ln kuk b C oe ln kln (sin v)k b C u c c c $ 201. ln y dy y xex e3x _ _ '1 x oe ln y ; dx oe dy y dy oe ex dx # oe " 2 x tanc" S 2 < b " # ax b 2 b bC '0 # # ' x c x b 2 dx oe ' x dx 2 c ' b # # ax b 2 b # oe " 2 x tanc" S 2 < b " ax# b 2b # tan t # c ' tan t dt oe dx oe '0 xe # # $ 196. x cos dx 2 ' sincos c dx x oe c ' (sin x) a1xcdx xb oe c ' (sincos axcos xb oe c ' sindx oe c2' csc 2x dx x sin sin x) 2x # ; u oe sinc" x du oe dx -- 1cx # # 193. ' 4)cd)) # oe ' ^c1 b 4 4c) d) oe c ' d ) c ' d) )c# b' ' cos u du oe sin u b C oe sin asinc" xb b C oe x b C "c " oe c x tanc" x b ln kxk c " # ln a1 b x# b b C oe c tanx x b ln kxk c ln 1 b x# b C dx xb# oe ln kx b 1k c ln kx b 2k b C d) )b# oe c) c ln k) c 2k b ln k) b 2k b C tan t # c ln ksec tk b C 2x dx oe lim <c x e # # " ' tan x x dx oe c " tanc" x b ' x a1dx x b oe c x tanc" x b ' dx c ' 1xbdx x b x x dx 1bx ; dv oe dx x " ,voecx; # bC 2x " c4e 2x `b 0 570 204. Chapter 8 Techniques of Integration oe secc" kuk b C oe secc" k2x c 1k b C 205. ' eln x dx oe ' x dx oe 2 x$# b C 3 ' e) 3 b 4e) d); " sin 5t ' 1 b (cos dt 5t) 206. u oe 4e) du oe 4e) d) " 4 " " ' 3 b u du oe 4 2 (3 b u)$# b C oe 6 a3 b 4e) b$# b C 3 209. 210. x& 5x% 20x$ 60x# 120x 120 0 sin x b c cos x c c sin x b cos x c sin x b c cos x c c sin x ' x& sin x dx oe cx& cos x b 5x% sin x b 20x$ cos x c 60x# sin x c 120x cos x b 120 sin x b C 211. ' 1 bdrr ; # $ % u oe r dr du oe 2r -- % 2 ' 2ubdu oe ' ^2 c 1 b u du oe 2u c 2 ln k1 b uk b C oe 2r c 2 ln ^1 b r b C 1 u oe secc" ku b 1k b C oe secc" kln t b 1k b C t # 216. ' t(1 b ln t)dt t)(2 b ln t) ; " u oe lndtt (ln du oe # # # 215. ' 8 dm m49m c 4 oe 8 7 ' dm mm c ^ 2 7 oe 4 secc" 7m b C # du ' (1 b u)u(2 b u) oe ' (u b 1)dub 1) c 1 (u $# $# # 214. ' (t b 1) dt at b 2tb ;" u oe t# b 2t du oe 2(t b 1) dt " # ' udu oe " # 3u"$ b C oe 3 # # $ # $ 213. 8 y ' y (ydy 2) oe ' dy c ' 2ydy b ' 4ydy c ' (y dy 2) oe ln y b 2 b 2 c y2 b b y y # % 212. c 10x b c ' x 4x 10x20x 9 dx oe ' d ^xx c 10x b 99 oe ln kx% c 10x# b 9k b C c b # bC "$ at# b 2tb b) ' (27)3)b1 d) oe " 3 ' (27)3)b1 d(3) b 1) oe # 208. ' dv e2v c 1 ;" x oe ev dx oe ev dv ' dx x x c 1 oe secc" x b C oe secc" aev b b C " 3 ln 27 " 3 (27)3)b" b C oe # # 207. ;" u oe cos 5t c"' 5 du oe c5 sin 5t dt du 1bu " " oe c 5 tanc" u b C oe c 5 tanc" (cos 5t) b C S 27 27 < b C ln 3 1 bC # # # # ' dx (2x c 1)x c x oe' 2 dx (2x c 1)4x c 4x oe' 2 dx (2x c 1)(2x c 1) c 1 ;" u oe 2x c 1 du oe 2 dx ' du u u c 1 Chapter 8 Practice Exercises 217. If u oe 571 '0x 1 b (t c 1)% dt and dv oe 3(x c 1)# dx, then du oe 1 b (x c 1)% dx, and v oe (x c 1)$ so integration 1 x by parts '0 3(x c 1)# ''0 1 b (t c 1)% dt" dx oe '(x c 1)$ ! '0x 1 b (t c 1)% dt" " ! 1 $# " c '0 (x c 1)$ 1 b (x c 1)% dx oe 'c " a1 b (x c 1)% b " oe 86c " 6 oe Av(v c 1) av b 1b b B(v c 1) av b 1b b Cv# av# b 1b b (Dv b E) av# b (v c 1) v oe 0: c1 oe cB B oe 1; v oe 1: 4 oe 2C C oe 2; coefficient of v% : 0 oe A b C b D A b D oe c2; coefficient of v$ : 4 oe cA b B b E c D coefficient of v# : 0 oe A c B b C c E C c D oe 4 D oe c2 (summing with previous equation); coefficient of v: 1 oe cA b B A oe 0; in summary: A oe 0, B oe 1, C oe 2, D oe c2 and E oe 1 oe lim <ln (v c 1)# c b_ b_ 1 # # # " v b tanc" v c ln a1 b v# b` 2 b 1) oe lim 'ln S (b c b < c 1b " b b tanc" b" c ^ln 1 c " # b tanc" 2 c ln 5 oe ^0 c 0 b 1 c ^0 c # oe b ln (5) b " # c tanc" 2 219. u oe f(x), du oe f w (x) dx; dv oe dx, v oe x; oe ^ 31 b c # 220. dx '0a 1bx 1a # ' 3 2 2 f(x) dx oe cx f(x)d 3 22 c ' 3 2 2 xf w (x) dx oe < 3#1 f ^ 3#1 c 1 f ^ 1 ` c ' 3 2 2 cos x dx # # _ oe ctanc" xd 0 oe tanc" a; 'a a therefore, tanc" a oe 1 # c tanc" a tanc" a oe CHAPTER 8 ADDITIONAL AND ADVANCED EXERCISES 2 sin x dx 1 c x 2x sin x dx 1 c x ' 2. " x 2x sin x dx 1 c x # asinc" xb dx oe x asinc" xb b 2 asinc" xb 1 c x# c 2x b C oe " x , " " " x(x b 1) oe x c x b 1 , " " " " x(x b 1)(x b 2) oe 2x c x b 1 b #(x b 2) , " " " " " x(x b 1)(x b 2)(x b 3) oe 6x c #(x b 1) b #(x b 2) c 6(x b 3) , " " " " " x(x b 1)(x b 2)(x b 3)(x b 4) oe #4x c 6(x b 1) b 4(x b #) c 6(x b 3) " x(x b 1)(x b 2) (x b m) oe # "c c' oe 2 asinc" xb 1 c x# c ' 2 dx oe 2 asinc" xb 1 c x# c 2x b C; therefore # oe! k 0 m (c") (k!)(m c k)!(x b k) ; k therefore ' # # u oe sinc" x, du oe dx 1 c x 2x ; dv oe c dx , v oe 21 c x# ; 1cx # "c ' asinc" xb dx oe x asinc" xb c ' # # # "c 1. u oe asinc" xb , du oe # ; dv oe dx, v oe x; ; # 1 1 c csin xd 3 2 2 oe 1 a3b c ab c c(c1) c 1d oe 1 a3b c ab b 2 # # dx 1 bx b oe lim ctanc" xd a oe lim atanc" b c tanc" ab oe b_ 1 4 a oe 1 since a 0. b_ b " 24(x b 4) the following pattern: dx x(x b 1)(x b 2) (x b m) 1 1 # # 1 1 1 1 # $ # _ '2 4v b v c 1 v (v c 1) av b 1b dv oe lim b_ 2 " '2b ^ v c 1 b vc# b 1 b v # # # # 1 1 # 218. 4v b v c 1 v (v c 1) av b 1b $ oe A v # b B v b C vc1 b Dv b E v b1 # 4v$ b v c 1 c 2v 1bv dv " # b tanc" 2 c ln 5 1 # c tanc" a; 572 Chapter 8 Techniques of Integration (c") oe ! ' (k!)(m c k)! ln kx b kk" b C k m k 0 ;" 4. ' sinc" y dy; dz oe z sin # # z oe y dy 2 y z " 6. u oe ln Sx b 1 b x< , du oe S x bdx 1 b x < S #x b oe bCoe # 4 oe " # ln St c 1 c t# < c 2x " # sinc" t b C uc1oe 2 3 2 3 sec ) du oe oe oe oe 4 3 2 3 sec ) tan ) d) " 3 " 3 4 ' S 3 sec ) b 1< (sec )) d) oe 4 3 tan ) b ln ksec ) b tan )k b C" oe " 3 " 3 3 (u c 1)# c 1 b 4 3u# c 6u c 1 b '2e2x c 2ex c ln u c 1 b (u c 1)# c 4 b SC" b 3 " 3 " b ln ex c 1 b e2x c 2ex c 3 " b C # # # # oe " 16 b 2 c 2 ' ' x 2x2x 2 2 b (x b 1) b 1 c x 2x2x 2b# b (x c 1) b 1 " dx b b c # # # # # % 9. ' x " 4 dx oe ' b " ax b 2b c 4x dx oe ' " ax b 2x b 2b ax c 2x b 2b dx # # 8. a ' 2e 3e2x c ex b dx c 6ex c 1 ;" u oe ex du oe ex dx ' (2u c 1) du 3u c 6u c 1 # # # oe " # " " du " ' u du 1 c # ' u du 1 c # ' uu b 1 oe # ln u c 1 c b u b1 c " # tanc" u b C oe oe " 3 ' (2u c 1) du (u c 1) c 4 3 4 3 ; " ' sec# ) d) b ' sec ) d) 3 3 # " 3 " 3 ln ln # d) oe du u b1 " # " ln tan ) c " c # ) b C sec ) (u c 1) b 3 (u c 1)# c 1 b C" 4 3 # < # # 7. ' dt t c 1 c t t oe sin ) ;" dt oe cos ) d) d) ' sincosc) cos ) oe ' tan d))c 1 ; du oe sec# ) d) ) u oe tan ) # # ' ln Sx b 1 b x< dx oe x ln Sx b 1 b x< c 2 # # tan ) c ln ksec ) b tan )k # 2x b x c ln 2x b 1 b 2x b x # " x b " oe " sec ) # # dx oe " sec ) tan ) d) -- # " 4 sec " ' (sec ) c^1)tan ))tan ) d) oe # ' asec# ) c sec )b d) bC x b x c ln 2x b 1 b 2x b x " # # " ' ln Sx b 1 b x< dx oe x ln Sx b 1 b x< c " ' xxdx b x ; " ' # # 1 # # # 5. ) ) cos " ' 1 cdtan ) oe ' cos cos sin ) d) oe ' 12bcos #2) d) oe # ' (sec 2) b 1) d) oe ln ksec 2) b4tan 2)k b 2) b C )c ) # # oey "c # "c oe b C " < 2 1 b x oe dx 2 x 1 b x ; dv oe dx, v oe x; ; x dx ^x b c 4 "c z1 c z c sin z bC 4 y c y sin y sinc" y b # c b # "c # # oe x # "c sinc" x b -- ' 2z sinc" z dz; from Exercise 3, ' z sinc" z dz ' sinc" y dy oe y sinc" y b y 1 c y#c sin y # # # oe x # sinc" x c " sin# ) d) oe x sinc" # # x1 c x c sin x bC 4 ' ) x c " ^# c # sin 2) 4 bCoe x # sinc" x b sin ) cos ) c ) 4 bC bC bC ' (u c 1)duu b 1b a # # # # # ' x sinc" x dx oe x# sinc" x c ' x dx 2 1 c x # # 3. u oe sinc" x, du oe oe dx 1 c x ; dv oe x dx, v oe x # ; x oe sin ) dx oe cos ) d) ' x sinc" x dx oe x # sinc" x c ' sin ) cos ) d) 2 cos ) Chapter 8 Additional and Advanced Exercises # # 573 oe " 16 ln x x b 2x b 2 c 2x b # b " ctanc" (x b 1) b tanc" (x c 1)d b C 8 " xb1 x x '12 ln u du oe cu ln u c ud # oe (2 ln 2 c 2) c (ln 1 c 1) oe 2 ln 2 c 1 oe ln 4 c 1 " " n c k # 15. dy dx oe 2 csin td ! 16. dy dx 1% oe1 # % # oe ^c " b ln 3 c (0 b ln 1) oe ln 3 c # b 1 shell shell 17. V oe 'a 21 ^ radius S height < dx oe '0 21xy dx oe 61 '0 x# 1 c x dx; 1 0 oe c61 '1 ^u"# c 2u$# b u&# du 0 c61 '1 (1 c u)# u du uoe1cx du oe c dx x# oe (1 c u)# oe c61 < 2 u$# c 4 u&# b 2 u(# ` " oe 61 ^ 2 c 3 5 7 3 84 16 ^ 70 c105b 30 oe 61 ^ 105 oe 321 oe 61 35 ! # 2 1cx 0 " # 4 5 b 2 7 b oe '0 S " c x < dx oe '0 ^c1 b 1 x # # 1 2 1 2 dx oe ' 1 2 ^c1 b " 1bx # # # # # # # oe 1 b S dy < oe dx # # c2x 1cx oe 1 b 2x b x a1 c x b b " 1cx b dx oe <cx b ln 1 c x ` ! 1 x # a1 c x b b 4x a1 c x b 1 x oe S 1 b x < ; L oe '0 c # 1 2 1 b S dy < dx dx "# # 1 1 oe cos 2x 1 b S dy < oe 1 b cos 2x oe 2 cos# x; L oe '0 dx # # oe '0 1 " 1 cx dx oe csinc" xd ! oe " 1 # oe k 0 k 0 1 c 'k S n <" " 4 1 b Scos 2t< dt oe 2 '0 # # # # oe # 14. n lim ! _ k 0 oe oe n lim ! S _ n n ck " < ^ n oe n lim ! _ c c n 1 n 1 oe n 13. n lim ! ln 1 b _ k 1 n k n " oe n lim ! ln ^1 b k ^ " ^ n oe '0 ln (1 b x) dx; " n _ n 1 k 1 n 1 b # " b " b # oe c b x! x! # lim lim x! Sc # 1 x 'x cos t dt oe t # b x! lim x 'x 1 cos t t ^ dt is an indeterminate 0 _ form and we apply l'Hopital's rule: c'1 x cos t t dt oe lim c S cos x < x < oe lim x! cos x oe 1 u oe 1 b x, du oe dx x oe 0 u oe 1, x oe 1 u oe 2 " " ^n # # b b # b # b 12. x! lim t! lim t S cos t < t # " 'x1 cos t dt; t S < oe lim t! c c 11. x lim _ ' xx sin t dt oe x lim_ cc cos td x x oe x lim_ cc cos x b cos (cx)d oe x lim_ ac cos x b cos xb oe x lim_ 0 oe 0 " cos t # # oe " 6 c ln x b " b x 1 " 1# 'ln x x cxb" bxb1 c 23 tanc" S 2x c 1 < c 23 tanc" S 2x b 1 <" b C 3 3 oe1 x! lim # # " # # # " # oe " 6 c ln x b 1 b x 1 " 1# " ' " x 2x cb 1 c ^ cx 3 xc b 3 4 c # # ' 10. " ' x " 1 dx oe 6 ' c " ^xc1 c b xc2 x cxb1 c xb2 x bxb1 2x b 1 x bxb1 dx c 3 ^x b b 3 4 dx 'x1 cos t dt diverges since '01 dt diverges; thus t t 4 cos# t dt 574 Chapter 8 Techniques of Integration b 4 % x oe 1 <ln 5 c x c 5 ` " oe 1 ^ln 4 c 5 c 1 ^ln x 4 oe 151 4 b 21 ln 4 shell shell 19. V oe 'a 21 ^ radius S height < dx oe '0 21xex dx b 1 oe 21 cxex c ex d " oe 21 ! 20. V oe '0 ln 2 21(ln 2 c x) aex c 1b dx c(ln 2) ex c ln 2 c xex b xd dx # oe 21 '0 ln 2 oe 21 '(ln 2) ex c (ln 2)x c xex b ex b oe 21 '2 ln 2 c (ln 2) c 2 ln 2 b 2 b oe 21 'c (ln#2) c ln 2 b 1" 21. (a) V oe '1 1 c1 c (ln x)# d dx e e # oe 1 cx c x(ln x)# d 1 b 21'1 ln x dx e (b) V oe '1 1(1 c ln x)# dx oe 1'1 c1 c 2 ln x b (ln x)# d dx e e (FORMULA 110) e oe 1 cx c x(ln x)# b 2(x ln x c x)d 1 e oe 1 ccx c x(ln x)# b 2x ln xd 1 oe 1 cce c e b 2e c (c1)d oe 1 e oe 1 cx c 2(x ln x c x) b x(ln x)# d 1 c 21'1 ln x dx e oe 1 cx c 2(x ln x c x) b x(ln x)# c 2(x ln x c x)d 1 e oe 1 c5x c 4x ln x b x(ln x)# d 1 oe 1 c(5e c 4e b e) c (5)d oe 1(2e c 5) e (b) V oe 1'0 aey c 1b# dy oe 1'0 ae2y c 2ey b 1b dy oe 1 < e# c 2ey b y` ! oe 1 'S e# c 2e b 1< c ^ " c 2" # # oe 1 S e# c 2e b 5 < oe # 23. (a) lim x ln x oe 0 # 1 ae c 4e b 5b # b b x! x! lim f(x) oe 0 oe f(0) f is continuous # 1 1 2y " # 22. (a) V oe 1 '0 <aey b# c 1` dy oe 1'0 ae2y c 1b dy oe 1 < e# c y` ! oe 1 ' e# c 1 c ^ " " oe # # 1 1 # # oe 1 '1 ^ dx b x 4 5 dx x b dx 5cx " 4 # # 18. V oe 'a 1y# dx oe 1 '1 25 dx x (5 c x) c 5 x # ln 2 " 0 (ln 2) # " c 21(ln 2 b 1) 2y " 1 ae c 3 b # Chapter 8 Additional and Advanced Exercises # u oe (ln x) du oe (2 ln x) dx # 2 x x # # # '2 x (b) V oe '0 1x (ln x) dx; # 1 OEb lim ' 3 (ln x) " b c 0 S 3 < (2 ln x) dv oe x dx ! voe x 3 $ 575 dx x 9 24. V oe '0 1(c ln x)# dx 1 25. M oe '1 ln x dx oe cx ln x c xd e oe (e c e) c (0 c 1) oe 1; 1 e Mx oe '1 (ln x) ^ ln#x dx oe e oe " # Scx(ln x)# d 1 c 2 '1 ln x dx< oe e e e # My oe '1 x ln x dx oe ' x e # 1 " 2x dx # 1 c x oe 2 'c 1 c x " oe 2; ! M 2 therefore, x oe My oe 1 and y oe 0 by symmetry 1 # My oe '0 28. y oe ln x 1 b S dx < oe 1 b x# S oe 21'c x1 b x# dy S oe 21'0 ey 1 b e2y dy; " dy # d 1 e tan 4 29. L oe 4 '0 1 b S dy < dx; x#$ b y#$ oe 1 y oe ^1 c x#$ dx 1 # # 1 oe 1 'e1 b e# b ln S b e be < 2b1 c 2" $# "# "c %1 oe 21 ^ " csec ) tan ) b ln ksec ) b tan )kd tan # e oe 1 'S1 b e# < e b ln 1 b e# b e" c 1 '2 1 b ln S2 b 1<" "c 1 u oe tan ) S oe 21'1 1 b u# du; " 21 ' du oe sec# ) d) e sec ) sec# ) d) dy dx oe c 3 ^1 c x#$ # # # oe S1 b e# c ln 1 b e e b " < c '2 c ln S1 b 2<" oe 1 b e# c ln S 1eb e b " < c 2 b ln S1 b 2< e e 1 4 4 "c 1 # tan e (sec )) atan ) b 1b tan ) tan "c "c 1 oe' d) oe ' e (tan ) sec ) b csc )) d) oe csec ) c ln kcsc ) b cot )kd tan4 1 dx; " # # e # " x e "c 27. L oe '1 1 b # 26. M oe '0 1 2 dx 1 c x oe 2 csinc" xd ! oe 1; dx oe '1 # therefore, x oe My M oe e b1 4 and y oe " x b 1 x # oe " # 'x# ln x c b oe c21 lim b! b oe 1 OE lim b! cx(ln x)# d b c 2'0 ln x dx9 1 1 cx ln x c xd 1 oe 21 b " # '1e (ln x)# dx " # e e ln x # "1 c " # e # x # " oe " # 'Se# c b b! x 9 b (e c 2); x dx " 4 '1 < b "" oe # Mx M ae# b 1b ; oe ec2 # tan x oe tan ) Loe' 4 # dx oe sec ) d) # $ oe 1 "^ 8 (ln 2)# c ^ 2 lim 3 3 ' x ln x c 3 " oe 1 ' 8(ln 2) c 3 2 16(ln 2) 9 b 16 27 " e sec )sec ) d) tan ) e $ $ b $ u oe ey du oe ey dy ^xc"$ ^ 2 3 576 Chapter 8 Techniques of Integration $# $# $# oe 41'0 ^1 c x#$ ^ x " dx; 1 1 u oe x#$ 3 $# '1 du oe c61'0 (1 c u)$# d(1 c u) 2 dx -- 4 # 1 0 (1 c u) du oe 3 x $" " oe c61 2 <(1 c u)&# ` ! oe 5 # 121 5 31. S dy < oe dx " 4x 1 dy dx oe ,," # x y oe x or y oe cx, 0 Y x Y 4 32. The integral ' 1 1 c x# dx is the area enclosed by the x-axis and the semicircle y oe 1 c x# . This area is half the circle's area, or 1 1 # and multiplying by 2 gives 1. The length of the circular arc y oe 1 c x# from x oe c1 to # 1 b b_ u oe ex " du oe ex dx b b_ oe ^c " b e! b ^0 b " oe 1 e e 34. u oe " 1by b b_ ! dy Y n lim _ oe " # b0oe " # ! oe " 2 <sinc" u ` # 0 2 oe " 2 sinc" 2 # oe " 2 ^1 4 # # $ # # 36. 1 6 oe sinc" " # 1 dx 4 c x 1 dx 4 c x c x oe 18 2 1 dx 4 c 2x oe " 2 b oe un 2 2 nb# bCoe 2 bCoe nb# " oe <sinc" x ` ! oe '0 # # # b ^ u n nb# '0 b Sx c a < n 2 bC '0 b # 35. u oe x# c a# du oe 2x dx; ' x Sx# c a# < n dx oe " ' ^u n du oe # " # ' un 2 du oe " S u b 1 < b C, n c2 # n 2 n 1 '0 2 du 4 c u c b # 0 Y n lim '0 _ 1 yn 1by '01 yn dy oe n lim_ ' nyb 1 " " oe n lim_ n 1 " n b1 oe 0 n lim '0 _ 1 # # # lim n_ '01 nyb y 1 c n 1 dy oe n lim OE' 1 y y " b '0 b _ n c dy , du oe c (1 b y) ; dv oe nyn # 1 dy, v oe yn ; 1 " yn 1by ^ c b a c oe a c_ <c " b e lim e b c_ ea `b lim c oe a c_ cce u d 1a b lim e c lim cc e u d e 1 c c lim ' e a c_ ea 1 u du b lim '1e b e u du b <ce eb b ca oe a c_ lim 'a0 e ex ex dx b b ca b ca _c b c a _c 33. (b) ex ex _ _ ' dx oe ' e ex ex dx lim (a) '0b e ex ex dx; b "` e dy9 oe " # b n lim _ # c circle's circumference. In conclusion, 2 ' 1 1 c x# dx oe ' 1 1 1 dx 1 c x . '01 yn 1by # c x oe 1 is L oe ' 1 1 b S dy < dx oe ' 1 1 b S cx < dx oe ' dx # # 1 1 1cx dx 1 c x oe " # (21) oe 1 since L is half the dy. Now, 0 Y yn 1by Y yn nyn 1 1by dy $# c $# 30. S oe 21 ' 1 f(x) 1 b cf w (x)d# dx; f(x) oe ^1 c x#$ 1 $" $# cf w (x)d# b 1 oe $" $# c c $# S dy < oe dx # 1cx x L oe 4 '0 1 b S 1 c x < dx oe 4'0 x 1 1 dx x oe 6 <x#$ ` ! oe 6 " x $# S oe 21 ' 1 ^1 c x#$ 1 " dx x c c c Chapter 8 Additional and Advanced Exercises # # 577 37. b_ b_ b_ " # b_ oe 1 if x 0 b_ b_ # " # p Y 1. 1 40. The area is given by the integral A oe '0 c dx xp ; oe c_, diverges; oe 1 c 0, converges; thus, p 1 for infinite area. 1 The volume of the solid of revolution about the x-axis is Vx oe 1'0 p " # volume for values of p satisfying 1 Y p 2, as described above. 41. e2x 2e2x 4e2x Ioe 42. e3x 3e3x 9e3x 3x b c b e2x 3 cos 3x 1 3 sin 3x c 1 cos 3x 9 2e2x 9 sin 3x b b c b cos 3x c 4 I 9 sin 4x " c 4 cos 4x " c 16 sin 4x 13 9 Ioe e2x 9 (3 sin 3x b 2 cos 3x) I oe e2x 13 (3 sin 3x b 2 cos 3x) b C I oe c e4 cos 4x b 3e3x 16 sin 4x c 9 16 I 25 16 Ioe e3x 16 (3 sin 4x c 4 cos 4x) I oe e3x 25 c converges if 2 p 1 p 2 (see Exercise 39). In conclusion, the curve y oe x p gives infinite area and finite (3 sin 4x c 4 cos 4x) b C _ The volume of the solid of revolution about the y-axis is Vy oe 1 '1 cR(y)d# dy oe 1'1 , and diverges if p " # . Thus, Vx is infinite whenever the area is infinite (p 1). _ b b b! b! c c p 1: A oe lim cx1 p d 1 oe " c lim b b b b! b! b1 c p 1: A oe lim cx1 p d 1 oe 1 c lim b b b p oe 1: A oe lim b! cln xd oe c lim 1 b b! ln b oe _, diverges; b1 p p c p " # for finite volume. In conclusion, the curve y oe x _ " about the x-axis is V oe '1 1 ^ xp dx oe 1 '1 _ _ 39. A oe '1 dx xp converges if p 1 and diverges if p Y 1. Thus, p Y 1 for infinite area. The volume of the solid of revolution dx x2p c xt b_ which converges if 2p 1 and diverges if 2p Y 1. Thus we want p gives infinite area and finite volume for values of p satisfying c c 38. G(x) oe lim # _ '1 b_ lim ln ^ x ax 1 c b " #x '0b e # ab b 1 b b a b_ b_ oe c_ the improper integral diverges if a " # " # ; in summary, the improper integral dx converges only when a oe and has the value c ln42 xb b dt oe lim <c " e xt ` 0 oe lim S 1 cxe < oe x 1c0 x oe " x if x 0 xG(x) oe x ^ " x dx x2p which converges if 2p 1 or dy y2 p c # oe " # ^ln 1 c " # ln 2 oe c ln42 ; if a " # : 0 Y lim ab b 1 b b b_ a lim (bb1)2a b b1 # b_ lim b_ # "ln oe lim (b b 1)2a 1 which # # integral diverges if a ; for a oe oe lim 1 b oe1 oe0 #" " # b b 1 " # : b lim b _ " b c ^ 'ln # # oe lim " b_ # ab b 1 b b a c ln 2a " ; lim ab b 1 b b b_ a lim b2a b_ b oe lim b2 a 1 2 oe _ if a " the improper ab b 1 b b # _ '1 ^ x ax 1 c b " #x dx oe lim '1b ^ x ax 1 c #"x dx oe b a lim < # ln ax# b 1b c " # " ln x` 1 oe lim ' # ln b ax b 1 b x a b " 1 c ln 2"# 578 43. Chapter 8 Techniques of Integration sin 3x 3 cos 3x b c b sin x ccos x csin x c9 sin 3x I oe c sin 3x cos x b 3 cos 3x sin x b 9I c8I oe c sin 3x cos x b 3 cos 3x sin x I oe sin 3x cos xc83 cos 3x sin x b C 44. cos 5x c sin 5x c25cos 5x Ioe 45. eax aeax a# eax ax b c b 5 16 sin 4x c " cos 4x 4 " c 16 sin 4 I oe c " cos 5x cos 4x c 4 " 9 sin 5x sin 4x b 25 16 9 I c 16 I oe c " cos 5x cos 4x c 4 5 16 sin 5x sin 4x (4 cos 5x cos 4x b 5 sin 5x sin 4x) b C b c b aeax b # sin bx c " cos bx b " c b sin bx # # # # # # 46. eax aeax a# eax Ioe eax b b c b # 47. ln (ax) " x b b I oe x ln (ax) c ' ^ " x dx oe x ln (ax) c x b C x 48. ln (ax) " x b b >(x b 1) oe '0 tx e t dt oe lim cctx e t d b b x '0 tx 1 e t dt oe lim ^c bb b 0x e! b x>(x) oe x>(x) 0 e b_ c c _ c c c c _ c (b) u oe tx , du oe xtx 1 dt; dv oe e t dt, v oe ce t ; x oe fixed positive real b_ b_ c c 49. (a) >(1) oe '0 e t dt oe lim c $ Ioe " 3 x$ ln (ax) c ' ^ " S x < dx oe x 3 _ # # Ioe eax a bb (a cos bx b b sin bx) b C 1 x x# " 3 x$ " 3 " x$ ln (ax) c 9 x$ b C '0b e t " dt oe lim cce t d b oe lim <c eb c (c1)` oe 0 b 1 oe 1 0 # # # # # # sin bx b # # Ioe eax a bb (a sin bx c b cos bx) b C cos bx " b sin bx " c b cos bx aeax b cos bx c a b I Sa bb b <I oe eax b (a cos bx b b sin bx) # # I oe c eb cos bx b sin bx c a b I Sa bb b <I oe eax b (a sin bx c b cos bx) b_ x b_ Chapter 8 Additional and Advanced Exercises (c) >(n b 1) oe n>(n) oe n!: n oe 0: >(0 b 1) oe >(1) oe 0!; n oe k: Assume >(k b 1) oe k! n oe k b 1: >(k b 1 b 1) oe (k b 1) >(k b 1) oe (k b 1)k! oe (k b 1)! Thus, >(n b 1) oe n>(n) oe n! for every positive integer n. 579 for some k 0; from part (b) induction hypothesis definition of factorial x n n 1 1 50. (a) >(x) ^ x 2x and n>(n) oe n! n! n ^ n 2n oe ^ n 2n1 e e e (b) n 10 20 30 40 50 60 n 10 ^ n n 2n1 e 3598695.619 2.4227868 , 10") 2.6451710 , 10$# 8.1421726 , 10%( 3.0363446 , 10'% 8.3094383 , 10)" ^ n n 2n1 e 3598695.619 calculator 3628800 2.432902 , 10") 2.652528 , 10$# 8.1591528 , 10%( 3.0414093 , 10'% 8.3209871 , 10)" ^ n n 2n1 e1 12n e 3628810.051 (c) calculator 3628800 580 Chapter 8 Techniques of Integration NOTES: ...
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This note was uploaded on 04/17/2008 for the course MA 113 taught by Professor Massman during the Spring '08 term at Rose-Hulman.

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