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Question Number 196928 by SANOGO last updated on 03/Sep/23

$$I_n={\int }_0^{\frac{\pi }{2}}{sin}^{n}xdx$$

Commented by Frix last updated on 03/Sep/23

$$\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}{\sin }^{n}xdx=\frac{1}{2}\mathrm{B}(\frac{1}{2},\frac{n+1}{2})=$$$$=\frac{\sqrt{\pi }\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}{2\mathrm{\Gamma }(\frac{n}{2}+1)}$$

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