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Question Number 177900 by yaslm last updated on 10/Oct/22

Answered by Ar Brandon last updated on 11/Oct/22

$$I_n={\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx$$$$I_{n+2}={\int }_0^{\frac{\pi }{2}}{\sin }^{n+2}xdx$$$$={\int }_0^{\frac{\pi }{2}}{\sin }^{n}x{\sin }^{2}xdx={\int }_0^{\frac{\pi }{2}}{\sin }^{n}x(1-{\cos }^{2}x)dx$$$$={\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx-{\int }_0^{\frac{\pi }{2}}{\sin }^{n}x{\cos }^{2}xdx$$$$=I_n-{\int }_0^{\frac{\pi }{2}}\left({\sin }^{n}x\cos x\right)\cos xdx$$$$=I_n-{\left[\frac{{\sin }^{n+1}x}{n+1}\cos x\right]}_0^{\frac{\pi }{2}}-\frac{1}{n+1}{\int }_0^{\frac{\pi }{2}}\left({\sin }^{n+1}x\right)\sin xdx$$$$=I_n-\frac{1}{n+1}{\int }_0^{\frac{\pi }{2}}{\sin }^{n+2}xdx=I_n-\frac{1}{n+1}{I}_{n+2}$$$$\Rightarrow (1+\frac{1}{n+1})I_{n+2}=I_n\Rightarrow \left(\frac{n+2}{n+1}\right)I_{n+2}=I_n\Rightarrow I_{n+2}=\frac{n+1}{n+2}{I}_n$$$$\Rightarrow I_{2n}=\frac{2n-1}{2n}{I}_{2n-2}=\frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\cdot \frac{2n-5}{2n-4}\cdot \cdot \cdot \frac{3}{4}\cdot \frac{1}{2}{I}_0$$$$I_0={\int }_0^{\frac{\pi }{2}}{\left(\sin x\right)}^{0}dx={\int }_0^{\frac{\pi }{2}}dx=\frac{\pi }{2}$$$$\Rightarrow I_{2n}=\frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\cdot \frac{2n-5}{2n-4}\cdot \cdot \cdot \frac{3}{4}\cdot \frac{1}{2}\cdot \frac{\pi }{2}$$

Commented by yaslm last updated on 11/Oct/22

thank you so much

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