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Question Number 142902 by greg_ed last updated on 07/Jun/21

I_{n} = \int_{0}^{_{} \frac{π}{2}} {(\sin x)}^{n} d x

with integration by parts, prove that :

I_{n + 2} = \frac{n + 1}{n + 2} . I_{n}

Answered by qaz last updated on 07/Jun/21

I_{n} = \int_{0}^{π / 2} \sin^{n} xdx

I_{n + 2} = \int_{0}^{π / 2} \sin^{n + 2} xdx

= \int_{0}^{π / 2} \sin^{n} x (1 - \cos^{2} x) dx

= I_{n} - \int_{0}^{π / 2} \sin^{n} xcos xd (\sin x)

= I_{n} - \frac{1}{n + 1} \int_{0}^{π / 2} \cos xd (\sin^{n + 1} x)

= I_{n} - \frac{1}{n + 1} (\sin^{n + 1} xcos x ∣_{0}^{π / 2} + \int_{0}^{π / 2} \sin^{n + 2} xdx)

= I_{n} - \frac{I_{n + 2}}{n + 1}

\Rightarrow I_{n + 2} = \frac{n + 1}{n + 2} \cdot I_{n}

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