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Question Number 63519 by Rio Michael last updated on 05/Jul/19

$$considerthegeneraldefiniteintergral$$$$I_n={\int }_0^{\frac{\pi }{2}}{sin}^{n}xdx$$$$a)provethatforn⩾2,{nI}_n=(n-1)I_{n-2}.$$$$b)Findthevaluesof\mathit{i}){\int }_0^{\frac{\pi }{2}}{sin}^{5}dx\mathit{i}\mathit{i}){\int }_0^{\frac{\pi }{2}}{sin}^{6}dx$$

Commented by Prithwish sen last updated on 05/Jul/19

$${\mathrm{I}}_{\mathrm{n}}={\int }_0^{\frac{\pi }{2}}{\sin }^{\mathrm{n}-1}\mathrm{x}\mathrm{sinxdx},\mathrm{using}\mathrm{by}\mathrm{parts}$$$$\text{Missing left or extra right}$$$$=(\mathrm{n}-1){\mathrm{I}}_{\mathrm{n}-2}-(\mathrm{n}-1){\mathrm{I}}_{\mathrm{n}}$$$${\mathrm{nI}}_{\mathrm{n}}=(\mathrm{n}-1){\mathrm{I}}_{\mathrm{n}-2}\mathbf{proved}$$$${\int }_0^{\frac{\pi }{2}}{\sin }^5\mathrm{x}\mathrm{dx}=\frac{(5-1)}{5}{\mathrm{I}}_3=\frac{4}{5}.\frac{(3-1)}{3}{\mathrm{I}}_{3-2}=\frac{4}{5}.\frac{2}{3}.\mathrm{I}$$$$=\frac{8}{15}{\int }_0^{\frac{\pi }{2}}\mathrm{sinx}\mathrm{dx}=-\frac{8}{15}{\left(\mathrm{cosx}\right)}_0^{\frac{\pi }{2}}=\frac{8}{15}$$$$\mathrm{similarly}\mathrm{you}\mathrm{can}\mathrm{deduce}\left(\mathrm{ii}\right)$$$$$$

Commented by Rio Michael last updated on 05/Jul/19

$$so\mathit{i}\mathit{i})becomes\frac{5\pi }{32}?$$

Commented by Prithwish sen last updated on 05/Jul/19

$$\mathrm{yes}$$

Commented by Rio Michael last updated on 05/Jul/19

$$okaysirthanks$$

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