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

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- Solution 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} .

Thus the recurrence relation is

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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