Exercise 1.2: Apply integration by parts a second time to get 272 2(1-2²) 2n x 2n 2(1-2²)-¹ sin(x2)dz = 2m ₁ [(1 – 2²)n-¹ – 22²(n − 1)(1 -- 2²)n-²] cos(xz)dz

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A: According to guideline we need to solve one question. Rest of the questions should be reposted.

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A: Please refer to the image below

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