B u 2 1 2x3 2 1x3 8 8 25 lim x 52x32 x 52

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Unformatted text preview: 9.4 310 1 dθ = cos2 θ 13. x = sin θ, dx = cos θ dθ, sec2 θ dθ = tan θ + C = x/ 1 − x2 + C √ 1 1 x2 + 25 sec θ +C 2 dθ = 25 csc θ cot θ dθ = − 25 csc θ + C = − 25x tan θ 1 14. x = 5 tan θ, dx = 5 sec θ dθ, 25 2 sec θ dθ = ln | sec θ + tan θ| + C = ln x + 15. x = sec θ, dx = sec θ tan θ dθ, x2 − 1 + C 16. 1 + 2x2 + x4 = (1 + x2 )2 , x = tan θ, dx = sec2 θ dθ, 1 dθ = sec2 θ = cos2 θ dθ = 1 2 (1 + cos 2θ)dθ = 1 1 θ + sin 2θ + C 2 4 1 1 x 1 θ + sin θ cos θ + C = tan−1 x + +C 2 2 2 2(1 + x2 ) 1 1 sec θ, dx = sec θ tan θ dθ, 3 3 17. x = 1 sec θ 2 dθ = 3 tan θ 1 3 1 csc θ cot θ dθ = − csc θ + C = −x/ 9x2 − 1 + C 3 18. x = 5 sec θ, dx = 5 sec θ tan θ dθ, sec3 θ dθ = 25 = 25 25 sec θ tan θ + ln | sec θ + tan θ| + C1 2 2 1 25 x x2 − 25 + ln |x + 2 2 x2 − 25| + C 19. ex = sin θ, ex dx = cos θ dθ, cos2 θ dθ = 20. u = sin θ, 1 2 √ 1 1 1 1 θ + sin 2θ + C = sin−1 (ex ) + ex 2 4 2 2 (1 + cos 2θ)dθ = 1 du = sin−1 2 − u2 sin θ √ 2 1 − e2x + C +C 21. x = 4 sin θ, dx = 4 cos θ dθ, π /2 1024 0 22. x = 1 24 1 1 sin3 θ cos2 θ dθ = 1024 − cos3 θ + cos5 θ 3 5 = 1024(1/3 − 1/5) = 2048/15 0 2 2 sin θ, dx = cos θ dθ, 3 3 π /6 0 1 1 dθ = 3θ cos 24 π /6 1 1 sec θ tan θ + ln | sec θ + tan θ| 48 48 0 √ √ √ √ 1 121 [(2/ 3)(1/ 3) + ln |2/ 3 + 1/ 3|] = + ln 3 = 48 48 3 2 π /3 π/4 √ 2 sec θ, dx = π /6 sec3 θ dθ = 23. x = sec θ, dx = sec θ tan θ dθ, 24. x = π /2 1 dθ = sec θ √ 2 sec θ tan θ dθ, 2 π /3 π /3 cos θ dθ = sin θ π/4 √ √ = ( 3 − 2)/2 π/4 π /4 π /4 tan2 θ dθ = 2 tan θ − 2θ 0 0 = 2 − π/2 0 311 Chapter 9 25. x = 1 9 √ 3 tan θ, dx = π /3 √ 3 sec2 θ dθ, 1 sec θ 4 dθ = 9 tan θ π/6 1 = 9 26. x = √ 3 3 √ π /3 1 cos3 θ 4 dθ = 9 sin θ π/6 √ 3/2 (u −4 −u 1/2 −2 π /3 π/6 1 1 − sin2 θ cos θ dθ = 4 9 sin θ 1 1 1 − 3+ )du = 9 3u u √ 3/2 1/2 √ 3/2 1/2 1 − u2 du (u = sin θ) u4 √ 10 3 + 18 = 243 √ 3 sec2 θ dθ, √ √ tan3 θ 3 π/3 3 3 π/3 dθ = sin θ dθ = 1 − cos2 θ sin θ dθ sec3 θ 30 30 √ √ π /3 1 1 1 1 3 3 3 − cos θ + cos θ −+ − −1 + = = 3 3 3 2 24 3 0 3 tan θ, dx = π /3 0 √ = 5 3/72 27. u = x2 + 4, du = 2x dx, 1 1 1 du = ln |u| + C = ln(x2 + 4) + C ; or x = 2 tan θ, dx = 2 sec2 θ dθ, u 2 2 1 2 tan θ dθ = ln | sec θ| + C1 = ln = √ x2 + 4 + C1 = ln(x2 + 4)1/2 − ln 2 + C1 2 1 ln(x2 + 4) + C with C = C1 − ln 2 2 1 1 x2 + 1 , 1 + (y )2 = 1 + 2 = , x x x2 √ 2 2 x2 + 1 x2 + 1 dx; x = tan θ, dx = sec2 θ dθ, dx = L= 2 x x 1 1 29. y = tan−1 (2) tan−1 (2) −1 tan (2) tan2 θ + 1 sec θ dθ = (sec θ tan θ + csc θ)dθ tan θ π/4 π/4 π/4 √ tan−1 (2) √ √ √ 51 − − = 5 + ln 2 + ln | 2 − 1| = sec θ + ln | csc θ − cot θ| 2 2 π/4 √ √ √ 2+2 2 √ = 5 − 2 + ln 1+ 5 L= sec3 θ dθ = tan θ 30. y = 2x, 1 + (y )2 = 1 + 4x2 , 1 1 + 4x2 dx; x = L= 0 1 L= 2 tan−1 2 1 1 tan θ, dx = sec2 θ dθ, 2 2 1 sec θ dθ = 2 3 0 1 1 sec θ tan θ + ln | sec θ + tan θ| 2 2 √ √ 1√ 1 1√ 1 = ( 5)(2) + ln | 5 + 2| = 5 + ln(2 + 5) 4 4 2 4 31. y = 2x, 1 + (y )2 = 1 + 4x2 , 1 x2 S = 2π 0 1 + 4x2 dx; x = 1 1 tan θ, dx = sec2 θ dθ, 2 2 tan−1 2 0 Exercise Set 9.4 tan−1 2 π S= 4 = 312 0 tan−1 2 π tan θ sec θ dθ = 4 2 3 0 tan−1 2 π (sec θ − 1) sec θ dθ = 4 2 (sec5 θ − sec3 θ)dθ 3 1 1 π1 sec3 θ tan θ − sec θ tan θ − ln | sec θ + tan θ| 44 8 8 tan−1 2 = 0 0 √ √ π [18 5 − ln(2 + 5)] 32 1 1 − y 2 dy ; y = sin θ, dy = cos θ dθ, y2 32. V = π 0 π /2 sin2 θ cos2 θ dθ = V =π 0 π /2 π 4 sin2 2θ dθ = 0 π 8 π /2 (1 − cos 4θ)dθ = 0 π 8 θ− 1 sin 4θ 4 π /2 = 0 du = u + C = sinh−1 (x/3) + C 33. (a) x = 3 sinh u, dx = 3 cosh u du, (b) x = 3 tan θ, dx = 3 sec2 θ dθ, sec θ dθ = ln | sec θ + tan θ| + C = ln( x2 + 9/3 + x/3) + C but sinh−1 (x/3) = ln(x/3 + x2 /9 + 1) = ln(x/3 + √ x2 + 9/3) so the results agree. (c) x = cosh u, dx = sinh u du, sinh2 u du = 1 2 (cosh 2u − 1)du = 1 1 sinh 2u − u + C 4 2 1 1 1 1 sinh u cosh u − u + C = x x2 − 1 − cosh−1 x + C 2 2 2 2 √ because cosh u = x, and sinh u = cosh2 u − 1 = x2 − 1 = 34. A = 4b a A=− 35. 37. 38. 39. 40. a a2 − x2 dx; x = a cos θ, dx = −a sin θ dθ, 0 π /2 0 4b a a2 sin2 θ dθ = 4ab π/2 0 1 1 dx = tan−1 2+9 (x − 2) 3 1 9 − (x − π /2 1)2 dx = sin−1 1 1 dx = 2+1 16(x + 1/2) 16 1 (x − 3)2 + 1 (1 − cos 2θ) dθ = πab sin2 θ dθ = 2ab 0 x−2 3 +C x−1 3 +C 36. 1 1 − (x − 1)2 dx = sin−1 (x − 1) + C 1 1 dx = tan−1 (4x + 2) + C 2 + 1/16 (x + 1/2) 4 dx = ln x − 3 + (x − 3)2 + 1 + C x dx, let u = x + 3, (x + 3)2 + 1 u−3 du = u2 + 1 = 3 u − u2 + 1 u2 + 1 du = 1 ln(u2 + 1) − 3 tan−1 u + C 2 1 ln(x2 + 6x + 10) − 3 tan−1 (x + 3) + C 2 π2 16 313 Chapter 9 4 − (x + 1)2 dx, let x + 1 = 2 sin θ, 41. cos2 θ dθ = 2θ + sin 2θ + C = 2θ + 2 sin θ cos θ + C 4 ex 42. (ex + 1/2)2 + 3/4 1 u2 + 3/4 1 + (x + 1) 3 − 2x − x2 + C 2 x+1 2 = 2 sin−1 dx, let u = ex + 1/2, √ du = sinh−1 (2u/ 3) + C = si...
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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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