U 2x du 2dx formula 83 2 u2 1 du 2 3x 3u sin2

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Unformatted text preview: 2 b b b eax sin bx dx = x2 dx 1 + x2 1 12 1 x tan−1 x − x + tan−1 x + C 2 2 2 ex sin x dx = −ex cos x + ex sin x − 2 12 1 x tan−1 x − 2 2 dx = x − tan−1 x + C1 so 19. u = ex , dv = sin x dx, du = ex dx, v = − cos x; For x tan−1 x dx = eax sin bx dx, eax (a sin bx − b cos bx) + C a2 + b2 e−3θ 22. From Exercise 21 with a = −3, b = 5, x = θ, answer = √ (−3 sin 5θ − 5 cos 5θ) + C 34 297 Chapter 9 23. u = sin(ln x), dv = dx, du = cos(ln x) 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, sin(ln x)dx = (x/2)[sin(ln x) − cos(ln x)] + C 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 = cos(ln x)dx, 1 x[cos(ln x) + sin(ln x)] + C 2 25. u = x, dv = sec2 x dx, du = dx, v = tan x; x sec2 x dx = x tan x − tan x dx = x tan x − sin 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 27. u = x2 , dv = xex dx, du = 2x dx, v = 2 x3 ex dx = 1 2 x2 xe − 2 28. u = xex , dv = 2 xex dx = 1 x2 e; 2 1 2 x2 1 x2 x e − e +C 2 2 1 1 ; dx, du = (x + 1)ex dx, v = − (x + 1)2 x+1 xex xex + dx = − (x + 1)2 x+1 ex dx = − xex ex + ex + C = +C x+1 x+1 Exercise Set 9.2 298 1 29. u = x, dv = e−5x dx, du = dx, v = − e−5x ; 5 1 0 1 xe−5x dx = − xe−5x 5 1 1 1 5 + 0 1 1 1 = − e−5 − e−5x 5 25 xe2x dx = 0 1 2x xe 2 2 − 0 2 1 2 1 2x e; 2 x2 ln x dx = 1 32. u = ln x, dv = e √ e 13 x ln x 3 1 e2x dx = e4 − e2x 4 0 e e 1 3 − 1 x2 dx = 1 e √ e e + √ e 1 dx x2 √ 1 1 1 = − + √ ln e − e x e e √ e 2 2 −2 −2 − −2 = 2 ln 5 − [x − 3 ln(x + 34. u = sin−1 x, dv = dx, du = √ = 2 √ 1 2 1 13 1 3 e − (e − 1) = (2e3 + 1)/9 3 9 θ, dv = dθ, du = x dx = 2 ln 5 + 2 ln 1 − x+3 2 3)]−2 2 −2 1− 3 dx x+3 = 2 ln 5 − (2 − 3 ln 5) + (−2 − 3 ln 1) = 5 ln 5 − 4 1 x 1 dx = sin−1 + 2 2 2 1−x 0 √ π 3 3 −1= + −1 4 12 2 π + 6 √ 1/2 − 0 sec−1 = 1 dx, v = x; 1 − x2 1/2 sin−1 x dx = x sin−1 x 0 4 e √ 1 1 1 1 3 e−4 =− + √ − +√ = e 2e e 2e e 2 ln(x + 3)dx = x ln(x + 3) √ 13 13 e− x 3 9 1 dx, v = x; x+3 33. u = ln(x + 3), dv = dx, du = 35. u = sec−1 0 1 = e4 − (e4 − 1) = (3e4 + 1)/4 4 1 1 1 dx, du = dx, v = − ; 2 x x x 1 ln x dx = − ln x 2 x x 1/2 2 1 1 dx, v = x3 ; x 3 31. u = ln x, dv = x2 dx, du = e 1 1 = − e−5 − (e−5 − 1) = (1 − 6e−5 )/25 5 25 0 30. u = x, dv = e2x dx, du = dx, v = 2 e−5x dx 0 √ 1/2 1 − x2 0 1 √ dθ, v = θ; 2θ θ − 1 √ √ 1 dθ = 4 sec−1 2 − 2 sec−1 2 − θ − 1 θ−1 2 2 √ 5π √ π π −2 − 3+1= − 3+1 =4 3 4 6 θdθ = θ sec−1 4 θ − 1 2 4 4 √ 2 299 Chapter 9 1 1 36. u = sec−1 x, dv = x dx, du = √ dx, v = x2 ; 2−1 2 xx 2 x sec−1 x dx = 1 = 2 12 x sec−1 x 2 − 1 2 1 2 √ 1 x dx x2 − 1 2 1 1 [(4)(π/3) − (1)(0)] − 2 2 x2 − 1 = 2π/3 − √ 3/2 1 1 37. u = x, dv = sin 4x dx, du = dx, v = − cos 4x; 4 π /2 0 π 38. π /2 1 x sin 4x dx = − x cos 4x 4 (x + x cos x)dx = 0 + 0 π 12 x 2 π /2 1 4 cos 4x dx = −π/8 + 0 π + x cos x dx = 0 0 π2 + 2 1 sin 4x 16 π /2 = −π/8 0 π x cos x dx; 0 u = x, dv = cos x dx, du = dx, v = sin x π π 0 3 √ √ x, dv = x tan−1 √ 0 √ 2 3/2 x tan−1 x 3 xdx = √ 2 3/2 x tan−1 x 3 = 3 − 1 3 − 1 40. u = ln(x2 + 1), dv = dx, du = 1 x dx 1+x 3 1− 1 1 dx 1+x 3 √ = (2 3π − π/2 − 2 + ln 2)/3 1 2x dx, v = x; +1 2 0 2 − ln(x2 + 1)dx = x ln(x2 + 1) 0 0 2x2 dx = 2 ln 5 − 2 2+1 x = 2 ln 5 − 2(x − tan−1 x) 2 0 2 1− 0 x2 1 +1 dx = 2 ln 5 − 4 + 2 tan−1 2 x, t2 = x, dx = 2t dt √ x e √ x e (b) 1 3 3 x2 2 (a) 1 3 √ 1 1 2 3/2 x tan−1 x − x + ln |1 + x| 3 3 3 = √ (x + x cos x)dx = π 2 /2 − 2 0 √ 1 2 xdx, du = √ dx, v = x3/2 ; 3 2 x(1 + x) 1 41. t = π = −2 so sin x dx = cos x 0 0 39. u = tan−1 π π − x cos x dx = x sin x cos cos dx = 2 tet dt; u = t, dv = et dt, du = dt, v = et , dx = 2tet − 2 √ x dx = 2 √ √ et dt = 2(t − 1)et + C = 2( x − 1)e x + C t cos t dt; u = t, dv = cos tdt, du = dt, v = sin t, √ x dx = 2t sin t − 2 √ √ √ sin tdt = 2t sin t + 2 cos t + C = 2 x sin x + 2 cos x + C Exercise Set 9.2 300 e e ln x dx = (x ln x − x) 43. (a) A = =1 1 1 e (ln x)2 dx = π (x(ln x)2 − 2x ln x + 2x) (b) V = π 1 π /2 (x − x sin x)dx = 44. A = 0 π /2 12 x 2 1 π /2 − x sin x dx = 0 0 = π (e − 2) π2 − (−x cos x + sin x) 8 π /2 = π 2 /8 − 1 0 π π x sin x dx = 2π (−x cos x + sin x) 45. V = 2π e 0 = 2π 2 0 π /2 π /2 46. V = 2π = π (π − 2) x cos x dx = 2π (cos x + x sin...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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