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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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- Fall '08
- SURGENT
- Calculus, Integration By Parts, Cos, dx