 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/17/c/solution

Two applications of integration by parts lower the exponent by two. First, the sine boundary term vanishes:

I_{n} = [x^{2 n} sin x]_{0}^{π} - 2 n \int_{0}^{π} x^{2 n - 1} sin x d x = - 2 n \int_{0}^{π} x^{2 n - 1} sin x d x .

For

n \geq 1

, integrate the last expression once more:

\int_{0}^{π} x^{2 n - 1} sin x d x = [- x^{2 n - 1} cos x]_{0}^{π} + (2 n - 1) \int_{0}^{π} x^{2 n - 2} cos x d x = π^{2 n - 1} + (2 n - 1) I_{n - 1} .

I_{n} = - 2 n π^{2 n - 1} - 2 n (2 n - 1) I_{n - 1}, n \geq 1.

The starting value is

I_{0} = \int_{0}^{π} cos x d x = 0

. Successive applications give

I_{1} = - 2 π

and

I_{2} = - 4 π^{3} + 24 π

, followed by

I_{3} = - 6 π^{5} + 120 π^{3} - 720 π .

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