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Question Number 137187 by mathocean1 last updated on 30/Mar/21

$$I_n={\int }_0^{\frac{\pi }{2}}{sin}^{n}xdx$$$$Writearelationbetween{I}_{n+2}and{I}_n.$$

Answered by Dwaipayan Shikari last updated on 30/Mar/21

$${\int }_0^{\frac{\pi }{2}}{sin}^{n}xdx={\int }_0^{\frac{\pi }{2}}{sin}^{2\left(\frac{n+1}{2}\right)-1}\left(x\right){cos}^{2\left(\frac{1}{2}\right)-1}\left(x\right)dx$$$$=\frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{2\mathrm{\Gamma }(\frac{n}{2}+1)}=J_n$$$$J_{n+2}=\frac{\mathrm{\Gamma }\left(\frac{n+3}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{2\mathrm{\Gamma }(\frac{n}{2}+2)}=\frac{\left(\frac{n+1}{2}\right)}{(\frac{n}{2}+1)}{J}_n$$

Answered by mathmax by abdo last updated on 30/Mar/21

$${\mathrm{I}}_{\mathrm{n}+2}={\int }_0^{\frac{\pi }{2}}{\sin }^{\mathrm{n}}\mathrm{x}{\sin }^2\mathrm{xdx}={\int }_0^{\frac{\pi }{2}}{\sin }^{\mathrm{n}}\mathrm{x}(1-{\cos }^2\mathrm{x})\mathrm{dx}$$$$={\int }_0^{\frac{\pi }{2}}{\sin }^{\mathrm{n}}\mathrm{xdx}-{\int }_0^{\frac{\pi }{2}}{\cos }^2\mathrm{x}{\sin }^{\mathrm{n}}\mathrm{x}\mathrm{dx}={\mathrm{I}}_{\mathrm{n}}-\mathrm{J}$$$$\mathrm{J}={\int }_0^{\frac{\pi }{2}}\mathrm{cosx}\left(\mathrm{cosx}{\sin }^{\mathrm{n}}\mathrm{x}\right)\mathrm{dx}{=}_{\mathrm{by}\mathrm{parts}}{\left[\frac{{\sin }^{\mathrm{n}+1}\mathrm{x}}{\mathrm{n}+1}\mathrm{cosx}\right]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}(-\mathrm{sinx})\frac{{\sin }^{\mathrm{n}+1}\mathrm{x}}{\mathrm{n}+1}\mathrm{dx}$$$$=\frac{1}{\mathrm{n}+1}{\int }_0^{\frac{\pi }{2}}{\sin }^{\mathrm{n}+2}\mathrm{xdx}\Rightarrow {\mathrm{I}}_{\mathrm{n}+2}={\mathrm{I}}_{\mathrm{n}}-\frac{1}{\mathrm{n}+1}{\mathrm{I}}_{\mathrm{n}+2}\Rightarrow$$$$(1+\frac{1}{\mathrm{n}+1}){\mathrm{I}}_{\mathrm{n}+2}={\mathrm{I}}_{\mathrm{n}}\Rightarrow \frac{\mathrm{n}+2}{\mathrm{n}+1}{\mathrm{I}}_{\mathrm{n}+2}={\mathrm{I}}_{\mathrm{n}}\Rightarrow {\mathrm{I}}_{\mathrm{n}+2}=\frac{\mathrm{n}+1}{\mathrm{n}+2}{\mathrm{I}}_{\mathrm{n}}$$

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