 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/5/solution

I = [- x^{2} cos x]_{0}^{π} + 2 \int_{0}^{π} x cos x d x = π^{2} + 2 ([x sin x]_{0}^{π} - \int_{0}^{π} sin x d x) .

The bracket containing

x sin x

vanishes, while the last definite integral is

2

. Therefore

I = π^{2} - 4.

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