271_HW7.1 - MAT 271 Summer 2011 Calculus II Scott Zinzer...

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MAT 271, Summer 2011 Calculus II, Scott Zinzer 7.1 HW 7.1. Integration by Parts. 14. Evaluate the integral Z θ sec 2 θ dθ Solution: We use integration by parts with u = θ , dv = sec 2 θ dθ . Then v = tan θ and du = . We now obtain Z θ sec 2 θ dθ = θ tan θ - Z tan θ dθ = θ tan θ + ln | cos θ | + C Recall that tan θ = sin θ cos θ so that the substitution method can be used to find R tan θ dθ . 21. Evaluate the integral Z x sin x cos x dx Solution: Recall that sin 2 x = 2 sin x cos x . Therefore, x sin x cos x = x 2 sin 2 x . We use integration by parts with u = x 2 and dv = sin 2 x . Then du = dx 2 and v = - 1 2 cos 2 x . We now obtain Z x sin x cos x dx = Z x 2 sin 2 x dx = - 1 4 x cos 2 x + Z 1 4 cos 2 x dx = - 1 4 x cos 2 x + 1 8 sin 2 x + C 26. Evaluate the integral Z x 2 ln 2 x dx Solution: We use integration by parts with u = ln 2 x and dv = x 2 dx . Then du = 2 ln x x dx and v = x 3 3 . We obtain Z x 2 ln 2 x dx = 1 3 x 3 ln 2 x - 2 3 Z x 2 ln x dx We use integration by parts again to find R x 2 ln x dx . This time, we let u = ln x and dv = x 2 dx . Then
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Unformatted text preview: x-1 3 Z x 2 dx = 1 3 x 3 ln x-1 9 x 3 + C Putting everything together, we get the following: Z x 2 ln 2 xdx = 1 3 x 3 ln 2 x-2 3 ± 1 3 x 3 ln x-1 9 x 3 + C ² = 1 3 x 3 ln 2 x-2 9 x 3 ln x + 2 27 x 3 + C MAT 271, Summer 2011 Calculus II, Scott Zinzer 7.1 HW 31. Evaluate the integral Z π/ 2 x cos 2 xdx Solution: We use integration by parts with u = x and dv = cos 2 xdx . Then du = dx and v = 1 2 sin 2 x . We obtain Z π/ 2 x cos 2 xdx = 1 2 x sin 2 x ± ± ± ± π/ 2-1 2 Z π/ 2 sin 2 xdx = 1 2 x sin 2 x ± ± ± ± π/ 2 + 1 4 cos 2 x ± ± ± ± π/ 2 = ²² π 4 sin( π ) ³-³ + ´´ 1 4 cos π µ-´ 1 4 cos 0 µµ = 0-1 4-1 4 =-1 2...
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