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

$${\mathrm{I}}_n={\int }_0^{{}_{}\frac{\pi }{2}}{\left(\sin x\right)}^{n}dx$$$$\mathbf{with}\mathbf{integration}\mathbf{by}\mathbf{parts},\mathbf{prove}\mathbf{that}:$$$${\mathrm{I}}_{n+2}=\frac{n+1}{n+2}.{\mathrm{I}}_n$$

Answered by qaz last updated on 07/Jun/21

$${\mathrm{I}}_{\mathrm{n}}={\int }_0^{\pi /2}\sin {}^{\mathrm{n}}\mathrm{xdx}$$$${\mathrm{I}}_{\mathrm{n}+2}={\int }_0^{\pi /2}\sin {}^{\mathrm{n}+2}\mathrm{xdx}$$$$={\int }_0^{\pi /2}\sin {}^{\mathrm{n}}\mathrm{x}(1-\cos {}^2\mathrm{x})\mathrm{dx}$$$$={\mathrm{I}}_{\mathrm{n}}-{\int }_0^{\pi /2}\sin {}^{\mathrm{n}}\mathrm{xcos}\mathrm{xd}\left(\sin \mathrm{x}\right)$$$$={\mathrm{I}}_{\mathrm{n}}-\frac{1}{\mathrm{n}+1}{\int }_0^{\pi /2}\cos \mathrm{xd}\left(\sin {}^{\mathrm{n}+1}\mathrm{x}\right)$$$$={\mathrm{I}}_{\mathrm{n}}-\frac{1}{\mathrm{n}+1}(\sin {}^{\mathrm{n}+1}\mathrm{xcos}\mathrm{x}{\mid }_0^{\pi /2}+{\int }_0^{\pi /2}\sin {}^{\mathrm{n}+2}\mathrm{xdx})$$$$={\mathrm{I}}_{\mathrm{n}}-\frac{{\mathrm{I}}_{\mathrm{n}+2}}{\mathrm{n}+1}$$$$\Rightarrow {\mathrm{I}}_{\mathrm{n}+2}=\frac{\mathrm{n}+1}{\mathrm{n}+2}\cdot {\mathrm{I}}_{\mathrm{n}}$$

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