Chapter 08

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Chapter 08

  1. 1. January 27, 2005 11:45 L24-CH08 Sheet number 1 Page number 337 black CHAPTER 8 Principles of Integral Valuation EXERCISE SET 8.1 1 1 1 1. u = 4 − 2x, du = −2dx, − u3 du = − u4 + C = − (4 − 2x)4 + C 2 8 8 3 √ 2. u = 4 + 2x, du = 2dx, u du = u3/2 + C = (4 + 2x)3/2 + C 2 1 1 1 3. u = x2 , du = 2xdx, sec2 u du = tan u + C = tan(x2 ) + C 2 2 2 4. u = x2 , du = 2xdx, 2 tan u du = −2 ln | cos u | + C = −2 ln | cos(x2 )| + C 1 du 1 1 5. u = 2 + cos 3x, du = −3 sin 3xdx, − = − ln |u| + C = − ln(2 + cos 3x) + C 3 u 3 3 2 2 1 du 1 1 2 6. u = x, du = dx, = tan−1 u + C = tan−1 x + C 3 3 6 1 + u2 6 6 3 7. u = ex , du = ex dx, sinh u du = cosh u + C = cosh ex + C 1 8. u = ln x, du = dx, sec u tan u du = sec u + C = sec(ln x) + C x 9. u = tan x, du = sec2 xdx, eu du = eu + C = etan x + C 1 du 1 1 10. u = x2 , du = 2xdx, √ = sin−1 u + C = sin−1 (x2 ) + C 2 1−u 2 2 2 1 1 6 1 11. u = cos 5x, du = −5 sin 5xdx, − u5 du = − u + C = − cos6 5x + C 5 30 30 √ du 1 + 1 + u2 1 + 1 + sin2 x 12. u = sin x, du = cos x dx, √ = − ln + C = − ln +C u u2 + 1 u sin x du 13. u = ex , du = ex dx, √ = ln u + u2 + 4 + C = ln ex + e2x + 4 + C 4 + u2 1 −1 14. u = tan−1 x, du = dx, eu du = eu + C = etan x +C 1 + x2 √ 1 √ 15. u = x − 1, du = √ dx, 2 eu du = 2eu + C = 2e x−1 +C 2 x−1 1 1 1 16. u = x2 + 2x, du = (2x + 2)dx, cot u du = ln | sin u| + C = ln sin |x2 + 2x| + C 2 2 2 √ 1 √ 17. u = x, du = √ dx, 2 cosh u du = 2 sinh u + C = 2 sinh x + C 2 x 337
  2. 2. January 27, 2005 11:45 L24-CH08 Sheet number 2 Page number 338 black 338 Chapter 8 dx du 1 1 18. u = ln x, du = , 2 =− +C =− +C x u u ln x √ 1 2 du 2 −u ln 3 2 −√x 19. u = x, du = √ dx, =2 e−u ln 3 du = − e +C =− 3 +C 2 x 3u ln 3 ln 3 20. u = sin θ, du = cos θdθ, sec u tan u du = sec u + C = sec(sin θ) + C 2 2 1 1 1 2 21. u = , du = − 2 dx, − csch2 u du = coth u + C = coth + C x x 2 2 2 x dx 22. √ = ln x + x2 − 4 + C x2 − 4 du 1 2+u 1 2 + e−x 23. u = e−x , du = −e−x dx, − = − ln + C = − ln +C 4−u 2 4 2−u 4 2 − e−x 1 24. u = ln x, du = dx, cos u du = sin u + C = sin(ln x) + C x ex dx du 25. u = ex , du = ex dx, √ = √ = sin−1 u + C = sin−1 ex + C 1 − e2x 1 − u2 1 26. u = x−1/2 , du = − dx, − 2 sinh u du = −2 cosh u + C = −2 cosh(x−1/2 ) + C 2x3/2 1 du 1 1 1 27. u = x2 , du = 2xdx, = sin u du = − cos u + C = − cos(x2 ) + C 2 csc u 2 2 2 2du 28. 2u = ex , 2du = ex dx, √ = sin−1 u + C = sin−1 (ex /2) + C 4 − 4u2 29. 4−x = e−x 2 2 ln 4 , u = −x2 ln 4, du = −2x ln 4 dx = −x ln 16 dx, 1 1 u 1 −x2 ln 4 1 −x2 − eu du = − e +C =− e +C =− 4 +C ln 16 ln 16 ln 16 ln 16 1 πx ln 2 1 30. 2πx = eπx ln 2 , 2πx dx = e +C = 2πx + C π ln 2 π ln 2 1 2 1 31. (a) u = sin x, du = cos x dx, u du = u + C = sin2 x + C 2 2 1 1 1 (b) sin x cos x dx = sin 2x dx = − cos 2x + C = − (cos2 x − sin2 x) + C 2 4 4 1 1 1 1 (c) − (cos2 x − sin2 x) + C = − (1 − sin2 x − sin2 x) + C = − + sin2 x + C, 4 4 4 2 and this is the same as the answer in part (a) except for the constants. 1 1 32. (a) sech 2x = = (now multiply top and bottom by sech2 x) cosh 2x cosh2 x + sinh2 x sech2 x = 1 + tanh2 x
  3. 3. January 27, 2005 11:45 L24-CH08 Sheet number 3 Page number 339 black Exercise Set 8.2 339 sech2 x (b) sech2x dx = dx = tan−1 (tanh x) + C, or, replacing 2x with x, 1 + tanh2 x sechx dx = tan−1 (tanh(x/2)) + C 1 2 2ex (c) sech x = = x = 2x cosh x e + e−x e +1 ex (d) sech x dx = 2 dx = 2 tan−1 (ex ) + C e2x+1 sec2 x 1 1 33. (a) = = tan x cos2 x tan x cos x sin x 1 1 1 sec2 x 1 (b) csc 2x = = = , so csc 2x dx = ln tan x + C sin 2x 2 sin x cos x 2 tan x 2 1 1 (c) sec x = = = csc(π/2 − x), so cos x sin(π/2 − x) 1 sec x dx = csc(π/2 − x) dx = − ln tan(π/2 − x) + C 2 EXERCISE SET 8.2 1. u = x, dv = e−2x dx, du = dx, v = − 1 e−2x ; 2 1 1 −2x 1 1 xe−2x dx = − xe−2x + e dx = − xe−2x − e−2x + C 2 2 2 4 1 3x 1 3x 1 1 3x 1 3x 2. u = x, dv = e3x dx, du = dx, v = e ; xe3x dx = xe − e3x dx = xe − e + C 3 3 3 3 9 3. u = x2 , dv = ex dx, du = 2x dx, v = ex ; x2 ex dx = x2 ex − 2 xex dx. For xex dx use u = x, dv = ex dx, du = dx, v = ex to get xex dx = xex − ex + C1 so x2 ex dx = x2 ex − 2xex + 2ex + C 1 1 4. u = x2 , dv = e−2x dx, du = 2x dx, v = − e−2x ; x2 e−2x dx = − x2 e−2x + xe−2x dx 2 2 For xe−2x dx use u = x, dv = e−2x dx to get 1 1 1 1 xe−2x dx = − xe−2x + e−2x dx = − xe−2x − e−2x + C 2 2 2 4 1 1 1 so x2 e−2x dx = − x2 e−2x − xe−2x − e−2x + C 2 2 4 1 5. u = x, dv = sin 3x dx, du = dx, v = − cos 3x; 3 1 1 1 1 x sin 3x dx = − x cos 3x + cos 3x dx = − x cos 3x + sin 3x + C 3 3 3 9
  4. 4. January 27, 2005 11:45 L24-CH08 Sheet number 4 Page number 340 black 340 Chapter 8 1 6. u = x, dv = cos 2x dx, du = dx, v = sin 2x; 2 1 1 1 1 x cos 2x dx = x sin 2x − sin 2x dx = x sin 2x + cos 2x + C 2 2 2 4 7. u = x2 , dv = cos x dx, du = 2x dx, v = sin x; x2 cos x dx = x2 sin x − 2 x sin x dx For x sin x dx use u = x, dv = sin x dx to get x sin x dx = −x cos x + sin x + C1 so x2 cos x dx = x2 sin x + 2x cos x − 2 sin x + C 8. u = x2 , dv = sin x dx, du = 2x dx, v = − cos x; x2 sin x dx = −x2 cos x + 2 x cos x dx; for x cos x dx use u = x, dv = cos x dx to get x cos x dx = x sin x + cos x + C1 so x2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + C 1 1 1 2 1 1 2 1 9. u = ln x, dv = x dx, du = dx, v = x2 ; x ln x dx = x ln x − x dx = x ln x − x2 + C x 2 2 2 2 4 √ 1 2 10. u = ln x, dv = dx, v = x3/2 ; x dx, du = x 3 √ 2 3/2 2 2 4 x ln x dx = x ln x − x1/2 dx = x3/2 ln x − x3/2 + C 3 3 3 9 ln x 11. u = (ln x)2 , dv = dx, du = 2 dx, v = x; (ln x)2 dx = x(ln x)2 − 2 ln x dx. x Use u = ln x, dv = dx to get ln x dx = x ln x − dx = x ln x − x + C1 so (ln x)2 dx = x(ln x)2 − 2x ln x + 2x + C 1 1 √ ln x √ 1 √ √ 12. u = ln x, dv = √ dx, du = dx, v = 2 x; √ dx = 2 x ln x−2 √ dx = 2 x ln x−4 x+C x x x x 3 3x 13. u = ln(3x − 2), dv = dx, du = dx, v = x; ln(3x − 2)dx = x ln(3x − 2) − dx 3x − 2 3x − 2 3x 2 2 but dx = 1+ dx = x + ln(3x − 2) + C1 so 3x − 2 3x − 2 3 2 ln(3x − 2)dx = x ln(3x − 2) − x − ln(3x − 2) + C 3 2x x2 14. u = ln(x2 + 4), dv = dx, du = 2+4 dx, v = x; ln(x2 + 4)dx = x ln(x2 + 4) − 2 dx x x2 + 4 x2 4 x but dx = 1− dx = x − 2 tan−1 + C1 so x2 +4 x2 + 4 2 x ln(x2 + 4)dx = x ln(x2 + 4) − 2x + 4 tan−1 +C 2
  5. 5. January 27, 2005 11:45 L24-CH08 Sheet number 5 Page number 341 black Exercise Set 8.2 341 √ 15. u = sin−1 x, dv = dx, du = 1/ 1 − x2 dx, v = x; sin−1 x dx = x sin−1 x − x/ 1 − x2 dx = x sin−1 x + 1 − x2 + C 2 16. u = cos−1 (2x), dv = dx, du = − √ dx, v = x; 1 − 4x2 2x 1 cos−1 (2x)dx = x cos−1 (2x) + √ dx = x cos−1 (2x) − 1 − 4x2 + C 1 − 4x2 2 3 17. u = tan−1 (3x), dv = dx, du = dx, v = x; 1 + 9x2 3x 1 tan−1 (3x)dx = x tan−1 (3x) − dx = x tan−1 (3x) − ln(1 + 9x2 ) + C 1 + 9x2 6 1 1 1 1 x2 18. u = tan−1 x, dv = x dx, du = dx, v = x2 ; x tan−1 x dx = x2 tan−1 x − dx 1 + x2 2 2 2 1 + x2 x2 1 but dx = 1− dx = x − tan−1 x + C1 so 1 + x2 1 + x2 1 1 1 x tan−1 x dx = x2 tan−1 x − x + tan−1 x + C 2 2 2 19. u = ex , dv = sin x dx, du = ex dx, v = − cos x; ex sin x dx = −ex cos x + ex cos x dx. For ex cos x dx use u = ex , dv = cos x dx to get ex cos x = ex sin x − ex sin x dx so ex sin x dx = −ex cos x + ex sin x − ex sin x dx, 1 x 2 ex sin x dx = ex (sin x − cos x) + C1 , ex sin x dx = e (sin x − cos x) + C 2 1 20. u = e3x , dv = cos 2x dx, du = 3e3x dx, v = sin 2x; 2 1 3x 3 e3x cos 2x dx = e sin 2x − e3x sin 2x dx. Use u = e3x , dv = sin 2x dx to get 2 2 1 3 e3x sin 2x dx = − e3x cos 2x + e3x cos 2x dx, so 2 2 1 3 9 e3x cos 2x dx = e3x sin 2x + e3x cos 2x − e3x cos 2x dx, 2 4 4 13 1 1 3x e3x cos 2x dx = e3x (2 sin 2x + 3 cos 2x) + C1 , e3x cos 2x dx = e (2 sin 2x + 3 cos 2x) + C 4 4 13 1 21. u = eax , dv = sin bx dx, du = aeax dx, v = − cos bx (b = 0); b 1 ax a eax sin bx dx = − e cos bx + eax cos bx dx. Use u = eax , dv = cos bx dx to get b b 1 a eax cos bx dx = eax sin bx − eax sin bx dx so b b 1 a a2 eax sin bx dx = − eax cos bx + 2 eax sin bx − 2 eax sin bx dx, b b b eax eax sin bx dx = 2 (a sin bx − b cos bx) + C a + b2
  6. 6. January 27, 2005 11:45 L24-CH08 Sheet number 6 Page number 342 black 342 Chapter 8 e−3θ 22. From Exercise 21 with a = −3, b = 5, x = θ, answer = √ (−3 sin 5θ − 5 cos 5θ) + C 34 cos(ln x) 23. u = sin(ln x), dv = dx, du = dx, v = x; x sin(ln x)dx = x sin(ln x) − cos(ln x)dx. Use u = cos(ln x), dv = dx to get cos(ln x)dx = x cos(ln x) + sin(ln x)dx so sin(ln x)dx = x sin(ln x) − x cos(ln x) − sin(ln x)dx, 1 sin(ln x)dx = x[sin(ln x) − cos(ln x)] + C 2 1 24. u = cos(ln x), dv = dx, du = − sin(ln x)dx, v = x; x cos(ln x)dx = x cos(ln x) + sin(ln x)dx. Use u = sin(ln x), dv = dx to get sin(ln x)dx = x sin(ln x) − cos(ln x)dx so cos(ln x)dx = x cos(ln x) + x sin(ln x) − cos(ln x)dx, 1 cos(ln x)dx = x[cos(ln x) + sin(ln x)] + C 2 25. u = x, dv = sec2 x dx, du = dx, v = tan x; sin x x sec2 x dx = x tan x − tan x dx = x tan x − dx = x tan x + ln | cos x| + C cos x 26. u = x, dv = tan2 x dx = (sec2 x − 1)dx, du = dx, v = tan x − x; x tan2 x dx = x tan x − x2 − (tan x − x)dx 1 1 = x tan x − x2 + ln | cos x| + x2 + C = x tan x − x2 + ln | cos x| + C 2 2 2 1 x2 27. u = x2 , dv = xex dx, du = 2x dx, v = e ; 2 2 1 2 x2 2 1 2 x2 1 x2 x3 ex dx = x e − xex dx = x e − e +C 2 2 2 1 1 28. u = xex , dv = dx, du = (x + 1)ex dx, v = − ; (x + 1)2 x+1 xex xex xex ex dx = − + ex dx = − + ex + C = +C (x + 1)2 x+1 x+1 x+1 1 2x 29. u = x, dv = e2x dx, du = dx, v = e ; 2 2 2 2 2 1 2x 1 1 1 xe2x dx = xe − e2x dx = e4 − e2x = e4 − (e4 − 1) = (3e4 + 1)/4 0 2 0 2 0 4 0 4
  7. 7. January 27, 2005 11:45 L24-CH08 Sheet number 7 Page number 343 black Exercise Set 8.2 343 1 30. u = x, dv = e−5x dx, du = dx, v = − e−5x ; 5 1 1 1 1 1 xe−5x dx = − xe−5x + e−5x dx 0 5 0 5 0 1 1 1 1 1 = − e−5 − e−5x = − e−5 − (e−5 − 1) = (1 − 6e−5 )/25 5 25 0 5 25 1 1 31. u = ln x, dv = x2 dx, du = dx, v = x3 ; x 3 e e e e 1 3 1 1 3 1 3 1 3 1 3 x2 ln x dx = x ln x − x2 dx = e − x = e − (e − 1) = (2e3 + 1)/9 1 3 1 3 1 3 9 1 3 9 1 1 1 32. u = ln x, dv = 2 dx, du = dx, v = − ; x x x e e e ln x 1 1 √ 2 dx = − ln x √ + √ dx e x x e e x2 √ 1 1 √ 1 e 1 1 1 1 3 e−4 = − + √ ln e − √ =− + √ − +√ = e e x e e 2 e e e 2e 1 33. u = ln(x + 2), dv = dx, du = dx, v = x; x+2 1 1 1 1 x 2 ln(x + 2)dx = x ln(x + 2) − dx = ln 3 + ln 1 − 1− dx −1 −1 −1 x+2 −1 x+2 1 = ln 3 − [x − 2 ln(x + 2)] = ln 3 − (1 − 2 ln 3) + (−1 − 2 ln 1) = 3 ln 3 − 2 −1 1 34. u = sin−1 x, dv = dx, du = √ dx, v = x; 1 − x2 √ 3/2 √ 3/2 √ 3/2 √ √ √ 3/2 −1 −1 x 3 3 sin x dx = x sin x − √ dx = sin−1 + 1 − x2 0 0 0 1 − x2 2 2 0 √ √ 3 π 1 π 3 1 = + −1= − 2 3 2 6 2 √ 1 35. u = sec−1 θ, dv = dθ, du = √ dθ, v = θ; 2θ θ − 1 4 √ √1 4 1 √ 4 √ 4 sec−1 θdθ = θ sec−1 dθ = 4 sec−1 2 − 2 sec−1 2 − θ − 1 θ − √ 2 2 2 θ−1 2 2 π π √ 5π √ =4 −2 − 3+1= − 3+1 3 4 6 1 1 36. u = sec−1 x, dv = x dx, du = √ dx, v = x2 ; x x2−1 2 2 2 2 1 2 1 x x sec−1 x dx = x sec−1 x − √ dx 1 2 1 2 1 x2 − 1 1 1 2 √ = [(4)(π/3) − (1)(0)] − x2 − 1 = 2π/3 − 3/2 2 2 1
  8. 8. January 27, 2005 11:45 L24-CH08 Sheet number 8 Page number 344 black 344 Chapter 8 1 37. u = x, dv = sin 2x dx, du = dx, v = − cos 2x; 2 π π π π 1 1 1 x sin 2x dx = − x cos 2x + cos 2x dx = −π/2 + sin 2x = −π/2 0 2 0 2 0 4 0 π π π π 1 2 π2 38. (x + x cos x)dx = x + x cos x dx = + x cos x dx; 0 2 0 0 2 0 u = x, dv = cos x dx, du = dx, v = sin x π π π π π x cos x dx = x sin x − sin x dx = cos x = −2 so (x + x cos x)dx = π 2 /2 − 2 0 0 0 0 0 √ √ 1 2 39. u = tan−1 x, dv = xdx, du = √ dx, v = x3/2 ; 2 x(1 + x) 3 3 √ √ 2 3/2 √ 3 1 3 x x tan−1 xdx = x tan−1 x − dx 1 3 1 3 1 1+x 2 3/2 √ 3 1 3 1 = x tan−1 x − 1− dx 3 1 3 1 1+x 2 3/2 √ 1 1 3 √ = x tan−1 x − x + ln |1 + x| = (2 3π − π/2 − 2 + ln 2)/3 3 3 3 1 2x 40. u = ln(x2 + 1), dv = dx, du = dx, v = x; x2 + 1 2 2 2 2 2x2 1 ln(x2 + 1)dx = x ln(x2 + 1) − dx = 2 ln 5 − 2 1− dx 0 0 0 x2+1 0 x2 + 1 2 = 2 ln 5 − 2(x − tan−1 x) = 2 ln 5 − 4 + 2 tan−1 2 0 √ 41. t = x, t2 = x, dx = 2t dt √ x (a) e dx = 2 tet dt; u = t, dv = et dt, du = dt, v = et , √ √ √ e x dx = 2tet − 2 et dt = 2(t − 1)et + C = 2( x − 1)e x + C √ (b) cos x dx = 2 t cos t dt; u = t, dv = cos tdt, du = dt, v = sin t, √ √ √ √ cos x dx = 2t sin t − 2 sin tdt = 2t sin t + 2 cos t + C = 2 x sin x + 2 cos x + C
  9. 9. January 27, 2005 11:45 L24-CH08 Sheet number 9 Page number 345 black Exercise Set 8.2 345 42. Let q1 (x), q2 (x), q3 (x) denote successive antiderivatives of q(x), so that q3 (x) = q2 (x), q2 (x) = q1 (x), q1 (x) = q(x). Let p(x) = ax2 + bx + c. Repeated Repeated Differentiation Antidifferentiation ax2 + bx + c q(x) + 2ax + b q1 (x) − 2a q2 (x) + 0 q3 (x) Then p(x)q(x) dx = (ax2 + bx + c)q1 (x) − (2ax + b)q2 (x) + 2aq3 (x) + C. Check: d [(ax2 +bx + c)q1 (x) − (2ax + b)q2 (x) + 2aq3 (x)] dx = (2ax + b)q1 (x) + (ax2 + bx + c)q(x) − 2aq2 (x) − (2ax + b)q1 (x) + 2aq2 (x) = p(x)q(x) 43. Repeated Repeated Differentiation Antidifferentiation 3x2 − x + 2 e−x + 6x − 1 −e−x − 6 e−x + 0 −e−x (3x2 − x + 2)e−x = −(3x2 − x + 2)e−x − (6x − 1)e−x − 6e−x + C = −e−x [3x2 + 5x + 7] + C 44. Repeated Repeated Differentiation Antidifferentiation x2 + x + 1 sin x + 2x + 1 − cos x − 2 − sin x + 0 cos x (x2 + x + 1) sin x dx = −(x2 + x + 1) cos x + (2x + 1) sin x + 2 cos x + C = −(x2 + x − 1) cos x + (2x + 1) sin x + C
  10. 10. January 27, 2005 11:45 L24-CH08 Sheet number 10 Page number 346 black 346 Chapter 8 45. Repeated Repeated Differentiation Antidifferentiation 4x4 sin 2x + 1 16x3 − cos 2x 2 − 1 48x 2 − sin 2x + 4 1 96x cos 2x 8 − 1 96 sin 2x 16 + 1 0 − cos 2x 32 4x4 sin 2x dx = (−2x4 + 6x2 − 3) cos 2x + −(4x3 + 6x) sin 2x + C 46. Repeated Repeated Differentiation Antidifferentiation √ x3 2x + 1 + 1 3x 2 (2x + 1)3/2 3 − 1 6x (2x + 1)5/2 15 + 1 6 (2x + 1)7/2 105 − 1 0 (2x + 1)9/2 945 √ 1 1 2 2 x3 2x + 1 dx = x3 (2x + 1)3/2 − x2 (2x + 1)5/2 + x(2x + 1)7/2 − (2x + 1)9/2 + C 3 5 35 315 47. (a) We perform a single integration by parts: u = cos x, dv = sin x dx, du = − sin x dx, v = − cos x, sin x cos x dx = − cos2 x − sin x cos x dx. Thus 1 2 sin x cos x dx = − cos2 x + C, sin x cos x dx = − cos2 x + C 2 1 2 1 (b) u = sin x, du = cos x dx, sin x cos x dx = u du = u + C = sin2 x + C 2 2 x 48. (a) u = x2 , dv = √ , du = 2x dx, v = x2 + 1, x2 +1 1 x3 1 1 √ 2 1 1√ 2 √ dx = x2 x2 + 1 − 2x x2 + 1 dx = 2 − (x2 + 1)3/2 =− 2+ 0 x2 + 1 0 0 3 0 3 3
  11. 11. January 27, 2005 11:45 L24-CH08 Sheet number 11 Page number 347 black Exercise Set 8.2 347 √ √ 2 2 x 1 3 (b) u = x2 + 1, du = √ dx, (u − 1) du = 2 u −u x2 + 1 1 3 1 2√ √ 1 1√ 2 = 2− 2− +1=− 2+ . 3 3 3 3 e e 49. (a) A = ln x dx = (x ln x − x) =1 1 1 e e (b) V = π (ln x)2 dx = π (x(ln x)2 − 2x ln x + 2x) = π(e − 2) 1 1 π/2 π/2 π/2 π/2 1 2 π2 50. A = (x − x sin x)dx = x − x sin x dx = − (−x cos x + sin x) = π 2 /8 − 1 0 2 0 0 8 0 π π 51. V = 2π x sin x dx = 2π(−x cos x + sin x) = 2π 2 0 0 π/2 π/2 52. V = 2π x cos x dx = 2π(cos x + x sin x) = π(π − 2) 0 0 π 53. distance = t3 sin tdt; 0 Repeated Repeated Differentiation Antidifferentiation t3 sin t + 3t2 − cos t − 6t − sin t + 6 cos t − 0 sin t π π t3 sin t dx = [(−t3 cos t + 3t2 sin t + 6t cos t − 6 sin t)] = π 3 − 6π 0 0 1 54. u = 2t, dv = sin(kωt)dt, du = 2dt, v = − cos(kωt); the integrand is an even function of t so kω π/ω π/ω π/ω π/ω 2 1 t sin(kωt) dt = 2 t sin(kωt) dt = − t cos(kωt) +2 cos(kωt) dt −π/ω 0 kω 0 0 kω k+1 π/ω k+1 2π(−1) 2 2π(−1) = + sin(kωt) = kω 2 k2 ω2 0 kω 2 1 3 55. (a) sin4 x dx = − sin3 x cos x + sin2 x dx 4 4 1 3 1 1 = − sin3 x cos x + − sin x cos x + x + C 4 4 2 2 1 3 3 = − sin3 x cos x − sin x cos x + x + C 4 8 8
  12. 12. January 27, 2005 11:45 L24-CH08 Sheet number 12 Page number 348 black 348 Chapter 8 π/2 π/2 π/2 1 4 (b) sin5 x dx = − sin4 x cos x + sin3 x dx 0 5 0 5 0 π/2 π/2 4 1 2 = − sin2 x cos x + sin x dx 5 3 0 3 0 π/2 8 8 =− cos x = 15 0 15 1 4 1 4 1 2 56. (a) cos5 x dx = cos4 x sin x + cos3 x dx = cos4 x sin x + cos2 x sin x + sin x + C 5 5 5 5 3 3 1 4 8 = cos4 x sin x + cos2 x sin x + sin x + C 5 15 15 1 5 (b) cos6 x dx = cos5 x sin x + cos4 x dx 6 6 1 5 1 3 = cos5 x sin x + cos3 x sin x + cos2 x dx 6 6 4 4 1 5 5 1 1 = cos5 x sin x + cos3 x sin x + cos x sin x + x + C, 6 24 8 2 2 π/2 1 5 5 5 cos5 x sin x + cos3 x sin x + cos x sin x + x = 5π/32 6 24 16 16 0 57. u = sinn−1 x, dv = sin x dx, du = (n − 1) sinn−2 x cos x dx, v = − cos x; sinn x dx = − sinn−1 x cos x + (n − 1) sinn−2 x cos2 x dx = − sinn−1 x cos x + (n − 1) sinn−2 x (1 − sin2 x)dx = − sinn−1 x cos x + (n − 1) sinn−2 x dx − (n − 1) sinn x dx, n sinn x dx = − sinn−1 x cos x + (n − 1) sinn−2 x dx, 1 n−1 sinn x dx = − sinn−1 x cos x + sinn−2 x dx n n 58. (a) u = secn−2 x, dv = sec2 x dx, du = (n − 2) secn−2 x tan x dx, v = tan x; secn x dx = secn−2 x tan x − (n − 2) secn−2 x tan2 x dx = secn−2 x tan x − (n − 2) secn−2 x (sec2 x − 1)dx = secn−2 x tan x − (n − 2) secn x dx + (n − 2) secn−2 x dx, (n − 1) secn x dx = secn−2 x tan x + (n − 2) secn−2 x dx, 1 n−2 secn x dx = secn−2 x tan x + secn−2 x dx n−1 n−1
  13. 13. January 27, 2005 11:45 L24-CH08 Sheet number 13 Page number 349 black Exercise Set 8.2 349 (b) tann x dx = tann−2 x (sec2 x − 1) dx = tann−1 x sec2 x dx − tann−2 x dx 1 = tann−1 x − tann−2 x dx n−1 (c) u = xn , dv = ex dx, du = nxn−1 dx, v = ex ; xn ex dx = xn ex − n xn−1 ex dx 1 1 1 59. (a) tan4 x dx = tan3 x − tan2 x dx = tan3 x − tan x + dx = tan3 x − tan x + x + C 3 3 3 1 2 1 2 (b) sec4 x dx = sec2 x tan x + sec2 x dx = sec2 x tan x + tan x + C 3 3 3 3 (c) x3 ex dx = x3 ex − 3 x2 ex dx = x3 ex − 3 x2 ex − 2 xex dx = x3 ex − 3x2 ex + 6 xex − ex dx = x3 ex − 3x2 ex + 6xex − 6ex + C 60. (a) u = 3x, 1 1 1 2 u 2 x2 e3x dx = u2 eu du = u2 eu − 2 ueu du = u e − ueu − eu du 27 27 27 27 1 2 u 2 2 1 2 2 = u e − ueu + eu + C = x2 e3x − xe3x + e3x + C 27 27 27 3 9 27 √ (b) u = − x, 1 √ −1 xe− x dx = 2 u3 eu du, 0 0 u3 eu du = u3 eu − 3 u2 eu du = u3 eu − 3 u2 eu − 2 ueu du = u3 eu − 3u2 eu + 6 ueu − eu du = u3 eu − 3u2 eu + 6ueu − 6eu + C, −1 −1 2 u3 eu du = 2(u3 − 3u2 + 6u − 6)eu = 12 − 32e−1 0 0 61. u = x, dv = f (x)dx, du = dx, v = f (x); 1 1 1 x f (x)dx = xf (x) − f (x)dx −1 −1 −1 1 = f (1) + f (−1) − f (x) = f (1) + f (−1) − f (1) + f (−1) −1 62. (a) u dv = uv − v du = x(sin x + C1 ) + cos x − C1 x + C2 = x sin x + cos x + C2 ; the constant C1 cancels out and hence plays no role in the answer. (b) u(v + C1 ) − (v + C1 )du = uv + C1 u − v du − C1 u = uv − v du
  14. 14. January 27, 2005 11:45 L24-CH08 Sheet number 14 Page number 350 black 350 Chapter 8 dx 63. u = ln(x + 1), dv = dx, du = , v = x + 1; x+1 ln(x + 1) dx = u dv = uv − v du = (x + 1) ln(x + 1) − dx = (x + 1) ln(x + 1) − x + C 3dx 2 64. u = ln(3x − 2), dv = dx, du = ,v = x − ; 3x − 2 3 2 2 1 ln(3x − 2) dx = u dv = uv − v du = x− ln(3x − 2) − x− dx 3 3 x − 2/3 2 2 = x− ln(3x − 2) − x − +C 3 3 1 1 65. u = tan−1 x, dv = x dx, du = 2 dx, v = (x2 + 1) 1+x 2 1 2 1 x tan−1 x dx = u dv = uv − v du = (x + 1) tan−1 x − dx 2 2 1 2 1 = (x + 1) tan−1 x − x + C 2 2 1 66. u = , dv = x dx, du = − x(ln x)2 dx, 1 1 v = ln x ln x 1 1 dx = 1 + dx. x ln x x ln x This seems to imply that 1 = 0, but recall that both sides represent a function plus an arbitrary constant; these two arbitrary constants will take care of the 1. 67. (a) u = f (x), dv = dx, du = f (x), v = x; b b b b f (x) dx = xf (x) − xf (x) dx = bf (b) − af (a) − xf (x) dx a a a a (b) Substitute y = f (x), dy = f (x) dx, x = a when y = f (a), x = b when y = f (b), b f (b) f (b) xf (x) dx = x dy = f −1 (y) dy a f (a) f (a) y (c) From a = f −1 (α) and b = f −1 (β) we get b bf (b) − af (a) = βf −1 (β) − αf −1 (α); then A1 β β f (b) f −1 (x) dx = f −1 (y) dy = f −1 (y) dy, a A2 α α f (a) x which, by Part (b), yields a= f –1(a) b=f –1(b) β b f −1 (x) dx = bf (b) − af (a) − f (x) dx α a f −1 (β) −1 −1 = βf (β) − αf (α) − f (x) dx f −1 (α) β f −1 (β) Note from the figure that A1 = f −1 (x) dx, A2 = f (x) dx, and α f −1 (α) A1 + A2 = βf −1 (β) − αf −1 (α), a “picture proof”.
  15. 15. January 27, 2005 11:45 L24-CH08 Sheet number 15 Page number 351 black Exercise Set 8.3 351 68. (a) Use Exercise 67(c); 1/2 sin−1 (1/2) π/6 1 1 1 1 sin−1 x dx = sin−1 −0·sin−1 0− sin x dx = sin−1 − sin x dx 0 2 2 sin−1 (0) 2 2 0 (b) Use Exercise 67(b); e2 ln e2 2 2 ln x dx = e2 ln e2 − e ln e − f −1 (y) dy = 2e2 − e − ey dy = 2e2 − e − ex dx e ln e 1 1 EXERCISE SET 8.3 1 1. u = cos x, − u3 du = − cos4 x + C 4 1 1 2. u = sin 3x, u5 du = sin6 3x + C 3 18 1 1 1 3. sin2 5θ = (1 − cos 10θ) dθ = θ− sin 10θ + C 2 2 20 1 1 1 4. cos2 3x dx = (1 + cos 6x)dx = x+ sin 6x + C 2 2 12 1 1 5. sin3 aθ dθ = sin aθ(1 − cos2 aθ) dθ = − cos aθ − cos3 aθ + C (a = 0) a 3a 6. cos3 at dt = (1 − sin2 at) cos at dt 1 1 = cos at dt − sin2 at cos at dt = sin at − sin3 at + C (a = 0) a 3a 1 1 7. u = sin ax, u du = sin2 ax + C, a = 0 a 2a 8. sin3 x cos3 x dx = sin3 x(1 − sin2 x) cos x dx 1 1 = (sin3 x − sin5 x) cos x dx = sin4 x − sin6 x + C 4 6 9. sin2 t cos3 t dt = sin2 t(1 − sin2 t) cos t dt = (sin2 t − sin4 t) cos t dt 1 1 = sin3 t − sin5 t + C 3 5 10. sin3 x cos2 x dx = (1 − cos2 x) cos2 x sin x dx 1 1 = (cos2 x − cos4 x) sin x dx = − cos3 x + cos5 x + C 3 5 1 1 1 1 11. sin2 x cos2 x dx = sin2 2x dx = (1 − cos 4x)dx = x− sin 4x + C 4 8 8 32
  16. 16. January 27, 2005 11:45 L24-CH08 Sheet number 16 Page number 352 black 352 Chapter 8 1 1 12. sin2 x cos4 x dx = (1 − cos 2x)(1 + cos 2x)2 dx = (1 − cos2 2x)(1 + cos 2x)dx 8 8 1 1 1 1 = sin2 2x dx + sin2 2x cos 2x dx = (1 − cos 4x)dx + sin3 2x 8 8 16 48 1 1 1 = x− sin 4x + sin3 2x + C 16 64 48 1 1 1 13. sin 2x cos 3x dx = (sin 5x − sin x)dx = − cos 5x + cos x + C 2 10 2 1 1 1 14. sin 3θ cos 2θdθ = (sin 5θ + sin θ)dθ = − cos 5θ − cos θ + C 2 10 2 1 1 15. sin x cos(x/2)dx = [sin(3x/2) + sin(x/2)]dx = − cos(3x/2) − cos(x/2) + C 2 3 3 16. u = cos x, − u1/3 du = − cos4/3 x + C 4 π/2 π/2 17. cos3 x dx = (1 − sin2 x) cos x dx 0 0 π/2 1 2 = sin x − sin3 x = 3 0 3 π/2 π/2 π/2 1 1 18. sin2 (x/2) cos2 (x/2)dx = sin2 x dx = (1 − cos 2x)dx 0 4 0 8 0 π/2 1 1 = x− sin 2x = π/16 8 2 0 π/3 π/3 π/3 1 1 19. sin4 3x cos3 3x dx = sin4 3x(1 − sin2 3x) cos 3x dx = sin5 3x − sin7 3x =0 0 0 15 21 0 π π π 1 1 1 20. cos2 5θ dθ = (1 + cos 10θ)dθ = θ+ sin 10θ =π −π 2 −π 2 10 −π π/6 π/6 π/6 1 1 1 21. sin 4x cos 2x dx = (sin 2x + sin 6x)dx = − cos 2x − cos 6x 0 2 0 4 12 0 = [(−1/4)(1/2) − (1/12)(−1)] − [−1/4 − 1/12] = 7/24 2π 2π 2π 1 1 1 1 22. sin2 kx dx = (1 − cos 2kx)dx = x− sin 2kx =π− sin 4πk (k = 0) 0 2 0 2 2k 0 4k 1 1 23. tan(2x − 1) + C 24. − ln | cos 5x| + C 2 5 25. u = e−x , du = −e−x dx; − tan u du = ln | cos u| + C = ln | cos(e−x )| + C 1 1 26. ln | sin 3x| + C 27. ln | sec 4x + tan 4x| + C 3 4
  17. 17. January 27, 2005 11:45 L24-CH08 Sheet number 17 Page number 353 black Exercise Set 8.3 353 √ 1 √ √ 28. u = x, du = √ dx; 2 sec u du = 2 ln | sec u + tan u| + C = 2 ln sec x + tan x + C 2 x 1 29. u = tan x, u2 du = tan3 x + C 3 1 1 30. tan5 x(1 + tan2 x) sec2 x dx = (tan5 x + tan7 x) sec2 x dx = tan6 x + tan8 x + C 6 8 1 1 31. tan 4x(1 + tan2 4x) sec2 4x dx = (tan 4x + tan3 4x) sec2 4x dx = tan2 4x + tan4 4x + C 8 16 1 1 32. tan4 θ(1 + tan2 θ) sec2 θ dθ = tan5 θ + tan7 θ + C 5 7 1 1 33. sec4 x(sec2 x − 1) sec x tan x dx = (sec6 x − sec4 x) sec x tan x dx = sec7 x − sec5 x + C 7 5 1 2 34. (sec2 θ − 1)2 sec θ tan θdθ = (sec4 θ − 2 sec2 θ + 1) sec θ tan θdθ = sec5 θ − sec3 θ + sec θ + C 5 3 35. (sec2 x − 1)2 sec x dx = (sec5 x − 2 sec3 x + sec x)dx = sec5 x dx − 2 sec3 x dx + sec x dx 1 3 = sec3 x tan x + sec3 x dx − 2 sec3 x dx + ln | sec x + tan x| 4 4 1 5 1 1 = sec3 x tan x − sec x tan x + ln | sec x + tan x| + ln | sec x + tan x| + C 4 4 2 2 1 5 3 = sec3 x tan x − sec x tan x + ln | sec x + tan x| + C 4 8 8 36. [sec2 x − 1] sec3 x dx = [sec5 x − sec3 x]dx 1 3 = sec3 x tan x + sec3 x dx − sec3 x dx (equation (20)) 4 4 1 1 = sec3 x tan x − sec3 x dx 4 4 1 1 1 = sec3 x tan x − sec x tan x − ln | sec x + tan x| + C (equation (20), (22)) 4 8 8 1 1 37. sec2 t(sec t tan t)dt = sec3 t + C 38. sec4 x(sec x tan x)dx = sec5 x + C 3 5 1 39. sec4 x dx = (1 + tan2 x) sec2 x dx = (sec2 x + tan2 x sec2 x)dx = tan x + tan3 x + C 3 40. Using equation (20), 1 3 sec5 x dx = sec3 x tan x + sec3 x dx 4 4 1 3 3 = sec3 x tan x + sec x tan x + ln | sec x + tan x| + C 4 8 8
  18. 18. January 27, 2005 11:45 L24-CH08 Sheet number 18 Page number 354 black 354 Chapter 8 41. u = 4x, use equation (19) to get 1 1 1 1 1 tan3 u du = tan2 u + ln | cos u| + C = tan2 4x + ln | cos 4x| + C 4 4 2 8 4 1 42. Use equation (19) to get tan4 x dx = tan3 x − tan x + x + C 3 √ 2 2 43. tan x(1 + tan2 x) sec2 x dx = tan3/2 x + tan7/2 x + C 3 7 2 44. sec1/2 x(sec x tan x)dx = sec3/2 x + C 3 π/8 π/8 1 45. (sec2 2x − 1)dx = tan 2x − x = 1/2 − π/8 0 2 0 π/6 π/6 1 46. sec2 2θ(sec 2θ tan 2θ)dθ = sec3 2θ = (1/6)(2)3 − (1/6)(1) = 7/6 0 6 0 47. u = x/2, π/4 1 π/4 √ 2 tan5 u du = tan4 u − tan2 u − 2 ln | cos u| = 1/2 − 1 − 2 ln(1/ 2) = −1/2 + ln 2 0 2 0 1 π/4 1 π/4 √ 48. u = πx, sec u tan u du = sec u = ( 2 − 1)/π π 0 π 0 1 1 49. (csc2 x − 1) csc2 x(csc x cot x)dx = (csc4 x − csc2 x)(csc x cot x)dx = − csc5 x + csc3 x + C 5 3 cos2 3t 1 1 50. · dt = csc 3t cot 3t dt = − csc 3t + C sin2 3t cos 3t 3 cos x 1 51. (csc2 x − 1) cot x dx = csc x(csc x cot x)dx − dx = − csc2 x − ln | sin x| + C sin x 2 1 52. (cot2 x + 1) csc2 x dx = − cot3 x − cot x + C 3 2π 2π 1 53. (a) sin mx cos nx dx = [sin(m + n)x + sin(m − n)x]dx 0 2 0 2π cos(m + n)x cos(m − n)x = − − 2(m + n) 2(m − n) 0 2π 2π but cos(m + n)x = 0, cos(m − n)x = 0. 0 0 2π 1 2π (b) cos mx cos nx dx = [cos(m + n)x + cos(m − n)x]dx; 0 2 0 since m = n, evaluate sin at integer multiples of 2π to get 0. 2π 1 2π (c) sin mx sin nx dx = [cos(m − n)x − cos(m + n)x] dx; 0 2 0 since m = n, evaluate sin at integer multiples of 2π to get 0.

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