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Ism et chapter_6

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Ism et chapter_6

  1. 1. 1 6. sec 2t tan 2t dt = sec 2t + c 2 7. (x2 + 4)2 dx = (x4 + 8x2 + 16)dx x5 8 = + x3 + 16x + c 5 3Chapter 6 8. x(x2 + 4)2 dx = (x5 + 8x3 + 16x)dx x6 = + 2x4 + 8x2 + cIntegration 3 6 3 x 9. dx = tan−1 + cTechniques 16 + x2 2 4 1 4 10. 2 dx = tan−1 x + c 4 + 4x 2 1 11. √ dx 3 − 2x − x26.1 Review of Formulas 1 x+1 = dx = arcsin +c and Techniques 4 − (x + 1)2 2 1 ax x+1 1. eax dx = e + c, for a = 0. 12. √ dx a 3 − 2x − x2 1 −2(x + 1) =− dx 1 2 4 − (x + 1)2 2. cos(ax)dx = sin(ax) + c, for a = 0. a 1 = − · 2[4 − (x + 1)2 ]1/2 + C 2 1 1 1 3. √ dx = dx = − 4 − (x + 1)2 + c a2 − x2 x 2 a 1− a 4 x 1 13. dx Let u = , du = dx. 5 + 2x + x2 a a 1 x+1 1 =4 dx = 2 tan−1 +c = √ du = sin−1 (u) + c 4 + (x + 1)2 2 1 − u2 −1 x = sin + c, a > 0. 4x + 4 a 14. dx 5 + 2x + x2 b 2(x + 1) 4. √ dx =2 dx = 2 ln | 4 + (x + 1)2 | + c |x| x2 − a2 4 + (x + 1)2 b 1 4t = dx 15. dt x 2 a 5 + 2t + t2 |x| a −1 4t + 4 4 x 1 = dt − dt Let u = , du = dx and |au| = |x| . 5 + 2t + t2 5 + 2t + t2 a a t+1 b = 2 ln 4 + (t + 1) 2 − 2tan−1 +c = √ du 2 |au| u2 − 1 b 1 t+1 2 (t + 1) = √ du 16. dt = dt |a| |u| u2 − 1 t2 + 2t + 4 2 (t + 1) + 3 b 1 = sec−1 (u) + c 2 = ln (t + 1) + 3 + c |a| 2 b x = sec−1 + c, a > 0. 17. 1 e3−2x dx = − e3−2x + c |a| a 2 1 3 5. sin(6t)dt = − cos(6t) + c 18. 3e−6x dx = − e−6x + c 6 6 360
  2. 2. 6.1. REVIEW OF FORMULAS AND TECHNIQUES 361 2 −1/3 30. Let u = ex , du = ex dx19. Let u = 1 + x2/3 , du = x dx 3 ex 1 4 3 √ dx = √ du dx = 4 u−1 du 1 − e2x 1 − u2 x1/3 (1 + x2/3 ) 2 = sin−1 u + C = sin−1 ex + c 2/3 = 6 ln |u| + C = 6 ln |1 + x |+c 31. Let u = x2 , du = 2xdx 3 x 1 120. Let u = 1 + x3/4 , du = x−1/4 dx √ dx = √ du 4 1 − x4 2 1 − u2 2 2 dx = dx 1 1 x1/4 + x x1/4 (1 + x3/4 ) = sin−1 u + C = sin−1 x2 + c 4 8 2 2 =2 u−1 du = ln |u| + C 3 3 32. Let u = 1 − x4 , du = −4x3 dx 8 2x3 1 = ln |1 + x3/4 | + c √ dx = − u−1/2 du 3 1−x 4 2 √ 1 = −u1/2 + C = −(1 − x4 )1/2 + c21. Let u = x, du = √ dx √ 2 x sin x 1+x √ dx = 2 sin udu 33. dx x 1 + x2 √ 1 1 2x = −2 cos u + C = −2 cos x + c = dx + dx 1 + x2 2 1 + x2 1 1 122. Let u = , du = − 2 dx = tan−1 x + ln |1 + x2 | + c x x 2 cos(1/x) dx = − cos udu 1 x2 34. √ dx 1 x+x = − sin u + C = − sin + c x 1 = x−1/2 · dx23. Let u = sin x, du = cos xdx 1 + x1/2 π 0 = 2 ln | 1 + x1/2 | + c cos xesin x dx = eu du = 0 0 0 ln x2 1 2 35. dx = 2 ln x dx24. Let u = tan x, du = sec xdx x x π/2 1 1 sec2 xetan x dx = eu du Let u = ln x, du = dx. x 0 0 2 u 1 =2 u du = u2 + c = (ln x) + c =e =e−1 0 3 3 3 0 x3 2625. sec x tan xdx 36. e2 ln x dx = x2 dx = = −π/4 1 1 3 1 3 0 √ 4 = sec x =1− 2 √ −π/4 37. x x − 3dx 3 π/2 4 √ π/226. csc2 xdx = − cot x =1 = (x − 3 + 3) x − 3dx π/4 π/4 3 4 4 3 2 = (x − 3)3/2 dx + 3 (x − 3)1/2 dx27. Let u = x , du = 3x dx 3 3 x2 1 1 2 4 2 4 12 dx = du = (x − 3)5/2 + 3 · (x − 3)3/2 = 1 + x6 3 1 + u2 5 3 5 3 3 1 1 = tan−1 u + C = tan−1 x3 + c 1 3 3 38. x(x − 3)2 dx 0 x5 1 128. dx = ln(1 + x6 ) + c = (x3 − 6x2 + 9x)dx 1 + x6 6 0 1 1 x x4 9 1129. √ dx = sin−1 + c = − 2x3 + x2 = 4−x 2 2 4 2 0 4
  3. 3. 362 CHAPTER 6. INTEGRATION TECHNIQUES 4 2 x2 + 1 39. √ dx 47. f (x)dx 1 x 0 4 4 1 2 x x2 = x3/2 dx + x−1/2 dx = 2+1 dx + dx 1 1 0 x 1 x2 + 1 1 2 4 1 1 2 5/2 4 72 = ln |x2 + 1| + 1− 2 dx = x + 2x1/2 = 2 0 1 x +1 5 1 1 5 1 2 = ln 2 + (x − arctan x) 0 0 2 1 2 1 −x2 e−4 − 1 ln 2 π 40. xe−x dx = − e = = + 1 + − arctan 2 −2 2 −2 2 2 4 4x + 1 5 5 x 48. dx 41. dx = √ arctan √ + c 2x2 + 4x + 10 3 + x2 3 3 5 4x + 4 3 dx: N/A = 2 + 4x + 10 dx − 2 + 4x + 10 dx 3 + x3 2x 2x 3 1 = ln |2x2 + 4x + 10| − dx 1 2 (x + 1)2 + 4 42. sin(3x)dx = sin(3x)3dx 3 3 x+1 Let u = 3x, du = 3dx. = ln |2x2 + 4x + 10| − tan−1 +c 1 1 4 2 = (sin u)du = − cos u + c 3 3 1 1 49. dx = tan−1 (x) + c. = − cos(3x) + c. (1 + x2 ) 3 x 1 2x 2) dx = dx (1 + x 2 (1 + x2 ) sin3 xdx = (sin2 x) sin xdx 1 = ln 1 + x2 + c. 2 = (1 − cos2 x)sin xdx x2 x2 + 1 − 1 dx = dx Let u = cos x, du = − sin xdx. (1 + x2 ) (1 + x2 ) 2 = 1 − u2 (−du) = u2 du − du x +1 1 = 2 + 1) dx − dx (x (1 + x2 ) u3 cos3 x 1 = −u= − cos x. = dx − dx 3 3 (1 + x2 ) = x − tan−1 (x) + c. 43. ln xdx: N/A x3 1 x2 2) dx = 2xdx Substituting u = ln x, (1 + x 2 (1 + x2 ) ln x 1 dx = ln2 x + c Let u = x2 , du = 2xdx. 2x 4 1 u 1 u+1−1 = du = du 2 1+u 2 1+u 44. Substituting u = x4 1 u+1 1 = du − du x3 1 2 1+u 1+u dx = arctan x4 + c 1 + x8 4 1 1 = du − du x4 2 1+u dx: N/A 1 1 + x8 = (u − ln (1 + u)) + c 2 1 1 45. e−x dx: N/A 2 = x2 − ln 1 + x2 + c. 2 2 Substituting u = −x2 Hence we can generalize this as follows, 2 1 2 xn xe−x dx = − e−x + c 1 + x2 dx 2 1 xn−2 = xn−1 − dx 46. sec xdx: N/A n−1 1 + x2 x 1 1 sec2 xdx = tan x + c 50. dx = 2xdx 1 + x4 2 1 + x4
  4. 4. 6.2. INTEGRATION BY PARTS 363 Let u = x2 , du = 2xdx. 1 2x 1 2x xe2x dx =xe − e dx 1 1 1 2 2 = du = tan−1 (u) + c 1 1 2 1 + u2 2 = xe2x − e2x + c. 1 2 4 = tan−1 x2 + c. 2 4. Let u = ln x, dv = x dx x3 1 1 1 x2 4 dx = 4x3 dx du = dx and v = . 1+x 4 1 + x4 x 2 Let u = 1 + x4 , du = 4x3 . 1 1 x ln x dx = x2 ln x − x dx 1 1 1 2 2 = du = ln (u) + c 1 1 4 u 4 = x2 ln x − x2 + c. 1 2 4 = ln 1 + x4 + c. 4 5. Let u = ln x, dv = x2 dx x5 1 1 dx du = dx, v = x3 . 1 + x4 x 3 1 x4 2 1 3 1 3 1 = 2xdx x ln xdx = x ln x − x · dx 2 1 + x4 3 3 x Let u = x2 , du = 2xdx. 1 3 1 = x ln x − x2 dx 1 u2 1 u2 + 1 − 1 3 3 = du = du 1 1 2 1+u 2 2 1 + u2 = x3 ln x − x3 + c. 1 u2 + 1 1 3 9 = du − du 2 1 + u2 1 + u2 1 6. Let u = ln x, du = dx. 1 1 x = du − du ln x u2 1 2 1 + u2 dx = udu = + c = (ln x)2 + c. 1 x 2 2 = u − tan−1 (u) + c 2 1 2 7. Let u = x2 , dv = e−3x dx = x − tan−1 x2 + c. 1 2 du = 2xdx, v = − e−3x Hence we can generalize this as follows, 3 x4n+1 1 x2n−2 x4(n−1)+1 I = x2 e−3x dx dx = − dx 1 + x4 2 n−1 1 + x4 1 1 and = − x2 e−3x − − e−3x · 2xdx 3 3 x4n+3 1 x2n x4(n−1)+3 1 2 dx = − dx = − x2 e−3x + xe−3x dx 1+x 4 4 n 1 + x4 3 3 Let u = x, dv = e−3x dx 16.2 Integration by Parts du = dx, v = − e−3x 3 1 I = − x2 e−3x 1. Let u = x, dv = cos xdx 3 du = dx, v = sin x. 2 1 1 + − xe−3x − − e−3x dx x cos xdx = x sin x − sin xdx 3 3 3 1 2 2 = x sin x + cos x + c = − x2 e−3x − xe−3x + e−3x dx 3 9 9 1 2 2 −3x 2. Let u = x, dv = sin 4xdx = − x2 e−3x − xe−3x − e +c 1 3 9 27 du = dx, v = − cos 4x 4 8. Let u = x3 , du = 3x2 dx. 3 1 1 x sin 4x dx x2 ex dx = eu dx = eu + c 3 3 1 1 1 x3 = − x cos 4x − − cos 4x dx = e + c. 4 4 3 1 1 = − x cos 4x + sin 4x + c. 4 16 9. Let I = ex sin 4xdx 3. Let u = x, dv = e2x dx u = ex , dv = sin 4xdx 1 1 du = dx, v = e2x . du = ex dx, v = − cos 4x 2 4
  5. 5. 364 CHAPTER 6. INTEGRATION TECHNIQUES 1 1 1 I = − ex cos 4x − − cos 4x ex dx du = cos xdx v = − cos 2x 4 4 2 1 x 1 1 1 1 = − e cos 4x + x e cos 4xdx I = cos x sin 2x + − cos 2x sin x 4 4 2 2 2 1 Use integration by parts again, this time let − − cos 2x cos xdx u = ex , dv = cos 4xdx 2 1 1 1 1 du = ex dx, v = sin 4x = cos x sin 2x − cos 2x sin x + Idx 4 2 4 4 1 x I = − e cos 4x So, 4 3 1 1 1 1 x 1 I = cos x sin 2x − cos 2x sin x + c1 + e sin 4x − (sin 4x)ex dx 4 2 4 4 4 4 1 1 1 2 1 I = − ex cos 4x + ex sin 4x − I I = cos x sin 2x − cos 2x sin x + c 4 16 16 3 3 So, 17 1 1 12. Here we use the trigonometric identity: I = − ex cos 4x + ex sin 4x + c1 sin 2x = 2 sin x cos x. 16 4 16 4 1 We then make the substitution I = − ex cos 4x + ex sin 4x + c 17 17 u = sin x, du = cos x dx. 10. Let, u = e2x , dv = cos x dx so that, sin x sin 2x dx = 2 sin2 x cos x dx du = 2e2x dx and v = sin x. e2x cos x dx 2 3 2 = 2u2 du = u + c = sin3 x + c 3 3 = e2x sin x − 2 e2x sin x dx This integral can also be done by parts, twice. If this is done, an equivalent answer is ob- Let, u = e2x , dv = sin x dx so that, tained: du = 2e2x dx and v = − cos x. 1 2 cos x sin 2x − cos 2x sin x + c e2x sin x dx 3 3 13. Let u = x, dv = sec2 xdx 2x 2x = −e cos x + 2 e cos x dx du = dx, v = tan x e2x cos x dx x sec2 xdx = x tan x − tan xdx sin x = e2x sin x + 2e2x cos x − 4 e2x cos x dx = x tan x − dx cos x Now we notice that the integral on both of Let u = cos x, du = − sin xdx 1 these is the same, so we bring them to one side x sec2 xdx = x tan x + du of the equation. u = x tan x + ln |u| + c 5 e2x cos x dx = x tan x + ln |cos x| + c = e2x sin x + 2e2x cos x + c1 14. Let u = (ln x)2 , dv = dx ln x e2x cos x dx du = 2 dx, v = x x 1 2x 2 = e sin x + e2x cos x + c I = (ln x)2 dx 5 5 ln x = x(ln x)2 − x·2 dx 11. Let I = cos x cos 2xdx x and u = cos x, dv = cos 2xdx = x(ln x)2 − 2 ln xdx 1 du = sin xdx, v = sin 2x Integration by parts again, 2 1 1 1 u = ln x, dv = dxdu = dx, v = x I = cos x sin 2x − sin 2x(− sin x)dx x 2 2 1 2 1 1 I = x(ln x) − 2 x ln x − x · dx = cos x sin 2x + sin x sin 2xdx x 2 2 Let,u = sin x, dv = sin 2xdx = x(ln x)2 − 2x ln x + 2 dx
  6. 6. 6.2. INTEGRATION BY PARTS 365 = x(ln x)2 − 2x ln x + 2x + c 20. Let u = 2x, dv = cos x dx duπ= 2 dxandv = sin x. π 215. Let u = x2 , dv = xex dx so that, du = 2x dx 2x cos xdx = 2x sin x|0 − 2 π sin xdx 1 2 0 0 and v = ex (v is obtained using substitu- = (2x sin x + π 2 cos x)|0 = −4. 2 tion). 2 1 2 2 1 x3 ex dx = x2 ex − xex dx 21. x2 cos πxdx 2 0 1 2 1 2 = x2 ex − ex + c Let u = x2 , dv = cos πxdx, 2 2 sin πx du = 2xdx, v = . π x 1 1 116. Let u = x2 , dv = dx sin πx sin πx 3/2 x2 cos πxdx = x2 − 2xdx (4 + x2 ) 0 π 0 0 π 1 2 1 du = 2xdx, v = − √ = (0 − 0) − x sin (πx) dx 4 + x2 π 0 3 1 x x 2 3/2 dx = x2 3/2 dx =− x sin (πx) dx (4 + x2 ) (4 + x2 ) π 0 x2 1 Let u = x, dv = sin(πx)dx, = −√ + √ 2xdx cos(πx) 4+x 2 4 + x2 du = dx, v = − . x2 π 1 =− + 2 (4 + x2 ) + c. 2 (4 + x2 ) − xsin(πx)dx π 0 1 117. Let u = ln(sin x), dv = cos xdx 2 x cos(πx) cos(πx) =− − − − dx 1 π π 0 0 π du = · cos xdx, v = sin x 1 sin x 2 cos π 1 sin(πx) =− (− − 0) + I = cos x ln(sin x)dx π π π π 0 = sin x ln(sin x) 2 1 1 2 1 =− + (0 − 0) = − 2 − sin x · · cos xdx π π π π sin x 1 = sin x ln(sin x) − cos xdx 22. x2 e3x dx = sin x ln(sin x) − sin x + c 0 Let u = x2 , dv = e3x dx, e3x18. This is a substitution u = x2 . du = 2xdx, v = . 1 3 x sin x2 dx = sin udu 1 x2 e3x 1 1 3x e 2 x2 e3x dx = − 2xdx 1 1 3 0 3 = − cos u + c = − cos x2 + c. 0 0 2 2 1 3 2 1 3x = e −0 − xe dx.19. Let u = x, dv = sin 2xdx 3 3 0 1 Let u = x, dv = e3x dx, du = dx, v = − cos 2x e3x 2 dv = dx, v = . 1 3 x sin 2xdx e3 2 1 3x 0 − xe dx 1 1 1 1 3 3 0 = − x cos 2x − − cos 2x dx e3 2 e3x 1 1 3x e 2 0 0 2 = − x − dx 1 1 1 3 3 3 0 0 3 = − (1 cos 2 − 0 cos 0) + cos 2xdx 1 2 2 0 e3 2 e3 e3x 1 = − − dx 1 1 1 3 3 3 0 3 = − cos 2 + sin 2x 3 1 2 2 2 0 e 2 e3 e3x 1 1 = − − = − cos 2 + (sin 2 − sin 0) 3 3 3 9 0 2 4 e 3 2 e3 1 3 1 1 = − − e −1 = − cos 2 + sin 2 3 3 3 9 2 4
  7. 7. 366 CHAPTER 6. INTEGRATION TECHNIQUES e3 2e3 2 3 x cos (ax) sin (ax) = − + e −1 =− + + c, a = 0. 3 9 27 a a2 e3 2e3 2e3 2 5e3 2 = − + − = − 3 9 27 27 27 27 27. (xn ) (ln x) dx = (ln x) (xn ) dx 10 Let u = ln x, dv = xn dx, 23. ln 2xdx 1 xn+1 1 du = dx, v = . Let u = ln 2x, dv = dx x (n + 1) 1 du = dx, v = x. (ln x)(xn ) dx x 10 10 10 1 xn+1 xn+1 dx ln (2x)dx = x ln (2x)|1 − x dx = (ln x) − 1 1 x (n + 1) (n + 1) x 10 = (10 ln(20) − ln 2) − dx xn+1 (ln x) xn = − dx 1 10 (n + 1) (n + 1) = (10 ln(20) − ln 2) − [x]1 x n+1 (ln x) x n+1 = (10 ln(20) − ln 2) − (10 − 1) = − 2 + c, n = −1. (n + 1) (n + 1) = (10 ln(20) − ln 2) − 9. 24. Let, u = ln x, dv = x dx 28. (sin ax) (cos bx) dx 1 x2 du = dx, v = . x 2 Let u = sin ax, dv = (cos bx) dx 2 2 2 sin bx 1 1 du = a (cos ax) dx, v = . x ln xdx = x2 ln x − xdx b 1 2 1 1 2 2 1 2 1 3 sin ax cos bx dx = x ln x − x2 = 2 ln 2 − . 2 4 1 4 sin bx sin bx = (sin ax) − a (cos ax) dx b b 25. x2 eax dx (sin ax) (sin bx) a = − (cos ax) (sin bx) dx b b Let u = x2 , dv = eax dx, Let u = cos ax, dv = sin bxdx, eax cos bx du = 2xdx, v = . du = −a (sin ax) dx, v = − . aax b e eax x2 eax dx = x2 − 2xdx sin ax sin bx a a a − cos ax sin bx dx b b x2 eax 2 sin ax sin bx a − cos bx = − xeax dx. = − cos ax a a b b b Let u = x, dv = eax dx, eax − cos bx dv = dx, v = . − (− sin ax) adx a b 2 ax x e 2 sin ax sin bx a − cos ax cos bx − xeax dx = − a a b b b x2 eax 2 eax eax a = − x − dx − cos bx sin ax dx a a a a b x2 eax 2 xeax eax sin ax sin bx a cos ax cos bx = − − 2 +c = + a a a a b b2 a 2 x2 eax 2xeax 2eax + sin ax cos bx dx = − + 3 + c, a = 0. b a a2 a sin ax cos bx dx 26. x sin (ax) dx sin ax sin bx a cos ax cos bx = + Let u = x, dv = sin axdx, b b2 cos ax a 2 du = dx, v = − . + sin ax cos bx dx a b a 2 xsin (ax) dx sin ax cos bx dx − sin ax cos bx dx b − cos (ax) cos (ax) sin ax sin bx a cos ax cos bx =x − − dx = + a a b b2
  8. 8. 6.2. INTEGRATION BY PARTS 367 a2 1− sin ax cos bx dx + (n − 1) sinn xdx b2 sin ax sin bx a cos ax cos bx n sinn xdx = + b b2 = − sinn−1 x cos x sin ax cos bx dx − (n − 1) sinn−2 xdx b2 sin ax sin bx a cos ax cos bx = + 1 b2 − a2 b b2 sinn xdx = − sinn−1 x cos x n sin ax cos bx dx n−1 − sinn−2 xdx 1 n = 2 − a2 (b sin ax sin bx + a cos ax cos bx) , b a = 0 b = 0. 31. x3 ex dx = ex (x3 − 3x2 + 6x − 6) + c29. Letu = cosn−1 x, dv = cos xdx du = (n − 1)(cosn−2 x)(− sin x)dx, v = sin x 32. cos5 xdx cosn xdx 1 4 = cos4 sin x + cos3 xdx = sin x cos n−1 x 5 5 1 − (sin x)(n − 1)(cosn−2 x)(− sin x)dx = cos4 sin x 5 = sin x cosn−1 x 4 1 2 + cos2 x sin x + cos xdx 5 3 3 + (n − 1)(cosn−2 x)(sin2 x)dx 1 4 = cos4 sin x + cos2 x sin x = sin x cosn−1 x 5 15 8 + (n − 1)(cosn−2 x)(1 − cos2 x)dx + sin x + c 15 = sin x cosn−1 x + (n − 1)(cosn−2 x − cosn x)dx 33. cos3 xdx 1 2 Thus, cosn xdx = cos2 x sin x + cos xdx 3 3 1 2 = sin x cosn−1 x + (n − 1) cosn−2 xdx = cos2 x sin x + sin x + c 3 3 − (n − 1) cosn xdx. 34. sin4 xdx n cosn xdx = sin x cosn−1 x 1 = − sin3 x cos x + 3 sin2 xdx 4 4 + (n − 1) cosn−2 xdx 1 3 1 1 = − sin3 x cos x + x − sin 2x 4 4 2 4 cosn xdx 1 1 n−1 35. x4 ex dx = sin x cosn−1 x + cos n−2 xdx 0 n n 1 = ex (x4 − 4x3 + 12x2 − 24x + 24) 030. Let u = sinn−1 x, dv = sin x dx = 9e − 24 du = (n − 1) sinn−2 x cos x, v = − cos x. sinn xdx 36. Using the work done in Exercise 34, π/2 = − sinn−1 x cos x sin4 xdx 0 + (n − 1) cos2 x sinn−2 xdx 1 3 3 π/2 = − sin3 x cos x + x − sin 2x = − sinn−1 x cos x 4 8 16 0 3π + (n − 1) (1 − sin2 x) sinn−2 xdx = 16 = − sinn−1 x cos x π/2 − (n − 1) sinn−2 xdx 37. sin5 xdx 0
  9. 9. 368 CHAPTER 6. INTEGRATION TECHNIQUES π/2 π/2 1 4 π/2 3 = − sin4 x cos x + sin xdx cosm xdx 5 0 5 0 0 1 π/2 (n − 1)(n − 3)(n − 5) · · · 2 = − sin4 x cos x = . 5 n(n − 2)(n − 4) · · · 3 0 π/2 4 1 2 41. Let u = cos−1 x, dv = dx + − sin2 x cos x − cos x 1 5 3 3 0 du = − √ dx, v = x (Using Exercise 30) 1 − x2 1 π π cos−1 xdx =− sin4 cos − sin4 0 cos 0 I= 5 2 2 4 1 2 π π 2 π 1 + − sin cos − cos = x cos−1 x − x −√ dx 5 3 2 2 3 2 1 − x2 8 x = = x cos−1 x + dx √ 15 1 − x2 Substituting u = 1 − x2 , du = −2xdx 38. Here we will again use the work we did in Ex- 1 1 ercise 34. I = x cos−1 x + √ − du u 2 sin6 xdx 1 = x cos1 x − u−1/2 du 1 5 2 = − sin5 x cos x + sin4 xdx 1 6 6 = x cos−1 x − · 2u1/2 + c 1 2 = − sin5 x cos x = x cos−1 x − 1 − x2 + c 6 5 1 3 3 + − sin3 x cos x + x − sin 2x + c 42. Let u = tan−1 x, dv = dx 6 4 8 16 1 1 5 du = dx, v = x = − sin5 x cos x − sin3 x cos x 1 + x2 6 24 x 15 + x− 15 sin 2x + c I= tan−1 xdx = x tan−1 x − dx 48 96 1 + x2 Substituting u = 1 + x2 , We now just have to plug in the endpoints: 1 π/2 I = x tan−1 x − ln(1 + x2 ) + c. sin6 xdx 2 0 √ 1 1 5 43. Substituting u = x, du = √ dx = − sin5 x cos x − sin3 x cos x 2 x 6 24 √ π/2 I = sin xdx = 2 u sin udu 15 15 + x− sin 2x 48 96 0 = 2(−u cos u + sin u) + c √ √ √ 15π = 2(− x cos x + sin x) + c = 96 √ 44. Substituting w = x 39. m even : 1 1 π/2 dw = √ dx = dx sinm xdx 2 x 2w √ x 0 (m − 1)(m − 3) . . . 1 π I= e dx = 2wew dx = · Next, using integration by parts m(m − 2) . . . 2 2 m odd: u = 2w, dv = ew dw π/2 du = 2dw, v = ew sinm xdx 0 I = 2wew − 2 ew dw (m − 1)(m − 3) . . . 2 √ √ √ = m(m − 2) . . . 3 = 2wew − 2ew + c = 2 xe x − 2e x + c 40. m even: 45. Let u = sin(ln x), dv = dx π/2 dx cosm xdx du = cos(ln x) , v = x 0 x π(n − 1)(n − 3)(n − 5) · · · 1 I = sin(ln x)dx = 2n(n − 2)(n − 4) · · · 2 m odd: = x sin(ln x) − cos(ln x)dx
  10. 10. 6.2. INTEGRATION BY PARTS 369 Integration by parts again, = 3 u2 eu − 2 ueu du u = cos(ln x), dv = dx dx du = − sin(ln x) , v = x = 3u2 eu − 6 ueu − eu du x cos(ln x)dx = 3u2 eu − 6ueu + 6eu + c 8 √ 2 3 x Hence e dx = 3u2 eu du = x cos(ln x) + sin(ln x)dx 0 0 2 I = x sin(ln x) − x cos(ln x) − I = 3u e − 6ueu + 6e 2 u u 0 = 6e2 − 6 2I = x sin(ln x) − x cos(ln x) + c1 1 1 50. Let u = tan−1 x, dv = xdx I = x sin(ln x) − x cos(ln x) + c dx x2 2 2 du = , v= 1 + x2 246. Let u = 4 + x2 , du = 2xdx I= x tan−1 xdx I= x ln(4 + x2 )dx x2 1 x2 = tan−1 x − dx 1 1 2 2 1 + x2 = ln udu = (u ln u − u) + C x2 2 2 = tan−1 x 1 2 = [(4 + x ) ln(4 + x2 ) − 4 − x2 ] + c 2 1 1 2 − 1dx − dx 2 1 + x247. Let u = e2x , du = 2e2x dx x2 1 1 = tan−1 x − x − tan−1 x + C I = e6x sin(e2x )dx = u2 sin udu 2 2 2 −1 x 2 x 1 Let v = u2 , dw = sin udu = tan x − + tan−1 x + c 2 2 2 dv = 2udu, w = − cos u 1 1 Hence x tan−1 xdx I= −u2 cos u + 2 u cos udu 0 2 x2 x 1 1 π 1 1 = tan−1 x − + tan−1 x = − = − u2 cos u + u cos udu 2 2 2 0 4 2 2 1 2 51. n times. Each integration reduces the power of = − u cos u + (u sin u + cos u) + c 2 x by 1. 1 = − e4x cos(e2x ) + e2x sin(e2x ) 2 52. 1 time. The first integration by parts gets rid + cos(e2x ) + c of the ln x and turns the integrand into a sim- ple integral. See, for example, Problem 4. √ 1 −2/348. Let u = 3 x = x1/3 , du = x dx, 53. (a) As the given problem, x sin x2 dx can 3 be simplified by substituting x2 = u, we 3u2 du = dx can solve the example using substitution I = cos x1/3 dx = 3 u2 cos udu method. Let v = u2 , dw = cos udu (b) As the given integral, x2 sin x dx can not dv = 2udu, w = sin u be simplified by substitution method and I = 3 u2 sin u − 2 u sin udu can be solved using method of integration by parts. = 3u2 sin u − 6 u sin udu (c) As the integral, x ln x dx can not be sim- plified by substitution and can be solved = 3u2 sin u − 6 −u cos u + cos udu using the method of integration by parts. 3 ln x = 3u sin u + 6u√ u −√ sin u + c √ √ cos 6 (d) As the given problem, dx can be = 3x sin 3 x + 6 3 x cos 3 x − 6 sin 3 x + c x simplified by substituting , ln x = u we √ 1 −2/3 can solve the example by substitution49. Let u = 3 x = x1/3 , du = x dx, method. 3 3u2 du = dx √ 3 54. (a) As this integral, x3 e4x dx can not be I = e x dx = 3 u2 eu du simplified by substitution method and can
  11. 11. 370 CHAPTER 6. INTEGRATION TECHNIQUES be solved by using the method of integra- 59. tion by parts. e2x 4 4 2x (b) As the given problem, x3 ex dx can be x e /2 + simplified by substituting x4 = u, we can 4x3 e2x /4 − solve the example using the substitution 12x2 e2x /8 + 2x method. 24x e /16 − 4 24 e2x /32 + (c) As the given problem, x−2 e x dx can be 1 simplified by substituting = u, we can x4 e2x dx x solve the example using the substitution x4 3x2 3x 3 = − x3 + − + e2x + c method. 2 2 2 4 (d) As this integral, x2 e−4x dx can not be 60. simplified by substitution and can be cos 2x solved by using the method of integration x 5 sin 2x/2 + by parts. 5x4 − cos 2x/4 − 55. First column: each row is the derivative of the 20x3 − sin 2x/8 + previous row; Second column: each row is the 60x2 cos 2x/16 − antiderivative of the previous row. 120x sin 2x/32 + 120 − cos 2x/64 − 56. sin x 4 x5 cos 2xdx x − cos x + 4x3 − sin x − 1 5 5 = x sin 2x + x4 cos 2x 12x2 cos x + 2 4 24x sin x − 20 60 − x3 sin 2x − x2 cos 2x 24 − cos x + 8 16 120 120 x4 sin xdx + x sin 2x + cos 2x + c 32 64 = −x4 cos x + 4x3 sin x + 12x2 cos x 61. − 24x sin x − 24 cos x + c e−3x 3 −3x 57. x −e /3 + cos x 3x2 e−3x /9 − x4 sin x + 6x −e−3x /27 + 4x3 − cos x − 6 e−3x /81 − 12x2 − sin x + x3 e−3x dx 24x cos x − 24 sin x + x3 x2 2x 2 = − − − − e−3x + c 3 3 9 27 x4 cos xdx 62. = x4 sin x + 4x3 cos x − 12x2 sin x x2 − 24x cos x + 24 sin x + c ln x x3 /3 + −1 58. x x4 /12 + ex −x −2 x5 /60 + 4 x ex + The table will never terminate. 4x3 ex − 12x2 ex + 63. (a) Use the identity 24x ex − cos A cos B 1 24 ex + = [cos(A − B) + cos(A + B)] 2 x4 ex dx This identity gives π 4 3 2 = (x − 4x + 12x − 24x + 24)e + c x cos(mx) cos(nx)dx −π
  12. 12. 6.2. INTEGRATION BY PARTS 371 π nπ 1 1 = [cos((m − n)x) = cos2 udu −π 2 n −nπ + cos((m + n)x)]dx nπ 1 1 1 1 sin((m − n)x) = u + cos(2u) =π = n 2 4 −nπ 2 m−n π π And then sin2 (nx)dx sin((m + n)x) −π + π m+n −π = (1 − cos2 (nx))dx =0 −π π π It is important that m = n because oth- = dx − cos2 (nx)dx erwise cos((m − n)x) = cos 0 = 1 −π −π (b) Use the identity = 2π − π = π sin A sin B 1 65. The only mistake is the misunderstanding of = [cos(A − B) − cos(A + B)] 2 antiderivatives. In this problem, ex e−x dx This identity gives π is understood as a group of antiderivatives of sin(mx) sin(nx)dx ex e−x , not a fixed function. So the subtraction −π π 1 by ex e−x dx on both sides of = [cos((m − n)x) −π 2 ex e−x dx = −1 + ex e−x dx − cos((m + n)x)]dx 1 sin((m − n)x) does not make sense. = π √ π 2 m−n π 66. V = π (x sin x)2 dx = π x2 sin xdx sin((m + n)x) 0 0 − m+n −π Using integration by parts twice we get =0 x2 sin xdx It is important that m = n because oth- erwise cos((m − n)x) = cos 0 = 1 = −x2 cos x + 2 x cos xdx64. (a) Use the identity = −x2 cos x + 2(x sin x − sin xdx) cos A sin B = −x2 cos x + 2x sin x + 2 cos x + c 1 = [sin(B + A) − sin(B − A)] Hence, 2 π This identity gives V = (−x2 cos x + 2x sin x + 2 cos x) 0 π = π 2 − 4 ≈ 5.87 cos(mx) sin(nx) dx −π π 1 67. Let u = ln x, dv = ex dx = [sin((n + m)x) dx −π 2 du = , v = ex x − sin((n − m)x)] dx ex ex ln xdx = ex ln x − dx 1 cos((n + m)x) x = − ex 2 n+m ex ln xdx + dx = ex ln x + C π x cos((n − m)x) Hence, + n−m 1 −π ex ln x + dx = ex ln x + c =0 x (b) We have seen that 68. We can guess the formula: 1 1 cos2 xdx = x + cos(2x) + c ex (f (x) + f (x))dx = ex f (x) + c 2 4 Hence by letting u = nx: and prove it by taking the derivative: π cos2 (nx)dx d x (e f (x)) = ex f (x) + ex f (x) −π dx
  13. 13. 372 CHAPTER 6. INTEGRATION TECHNIQUES b = ex (f (x) + f (x)) + f (x) (b − x) dx a 1 b 69. Consider, f (x)g (x) dx Consider x sin (b − x) dx 0 0 Choose u = g (x) and dv = f (x)dx, b b so that du = g (x) dx and , v = f (x) . = (b − x) sin xdx = (sin x) (b − x) dx 0 0 Hence, we have Now, consider 1 g (x)f (x)dx f (x) = x − sin x ⇒ f (x) = 1 − cos x 0 and f (x) = sin x. 1 1 = g (x) f (x)|0 − f (x)g (x) dx Therefore, using 0 f (b) = f (a) + f (a) (b − a) = (g (1) f (1) − g (0) f (0)) b 1 + f (x) (b − x) dx, − g (x)f (x) dx a 0 we get From the given data. b − sin b = 0 − sin 0 + f (0) (b − 0) 1 b = (0 − 0) − g (x)f (x) dx. + (sin x) (b − x) dx 0 0 Choose, u = g (x) and dv = f (x)dx, b so that,du = g (x) dx and v = f (x) . ⇒ |sin b − b| = x sin (b − x) dx . 0 Hence, we have 1 Further, − g (x)f (x) dx b b 0 |sin b − b| = x sin (b − x) dx ≤ xdx , 1 0 0 1 =− g (x) f (x)|0 − f (x) g (x) dx as sin (b − x) ≤ 1. 0 = − {(g (1) f (1) − g (0) f (0) ) b2 Thus, |sin b − b| ≤ . 1 2 − f (x) g (x) dx Therefore the error in the approximation 0 1 From the given data. sin x ≈ x is at most x2 . 1 2 = − (0 − 0 ) − f (x) g (x) dx 1 0 6.3 Trigonometric = f (x) g (x) dx. 0 Techniques of 70. Consider, Integration b b f (x) (b − x) dx = (b − x) f (x) dx 1. Let u = sin x, du = cos xdx a a Choose u = (b − x) and dv = f (x) dx, cos x sin4 xdx = u4 du so that du = −dx and v = f (x) . 1 5 1 = u + c = sin5 x + c Hence, we have: 5 5 b (b − x)f (x) dx 2. Let u = sin x, du = cos xdx a b cos3 x sin4 xdx = (1 − u2 )u4 du b = (b − x) f (x)|a + f (x) dx a u5 u7 b = − +c 5 7 = (0 − [(b − a) f (a)]) + f (x) dx sin x sin7 x 5 a b = − +c = − [(b − a) f (a)] + f (x)|a 5 7 = − [(b − a) f (a)] + f (b) − f (a) 3. Let u = sin 2x, du = 2 cos 2xdx. b π/4 f (x) (b − x) dx cos 2xsin3 2xdx a 0 = − [(b − a) f (a)] + f (b) − f (a) 1 1 1 1 u4 1 = u3 du = = f (b) = f (a) + (b − a) f (a) 2 0 2 4 0 8

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