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Question Number 161617 by gbanda95 last updated on 20/Dec/21

Answered by mathmax by abdo last updated on 20/Dec/21

$${\mathrm{I}}_{\mathrm{n}+2}={\int }_0^{\frac{\pi }{2}}{\sin }^{\mathrm{n}+2}\mathrm{dx}={\int }_0^{\frac{\pi }{2}}(1-{\cos }^2\mathrm{x}){\sin }^{\mathrm{n}}\mathrm{x}\mathrm{dx}$$$$={\mathrm{I}}_{\mathrm{n}}-{\int }_0^{\frac{\pi }{2}}\mathrm{cosx}\left(\mathrm{cosx}{\sin }^{\mathrm{n}}\mathrm{x}\right)\mathrm{dx}(\mathrm{u}=\mathrm{cosx}\mathrm{and}{\mathrm{v}}^{{}^{\prime }}=\mathrm{cosx}{\sin }^{\mathrm{n}}\mathrm{x})$$$$\mathrm{so}{\int }_0^{\frac{\pi }{2}}\mathrm{cosx}\left(\mathrm{cosx}{\sin }^{\mathrm{n}}\mathrm{x}\right)\mathrm{dx}={\left[\mathrm{cosx}\frac{1}{\mathrm{n}+1}{\sin }^{\mathrm{n}+1}\mathrm{x}\right]}_0^{\frac{\pi }{2}}$$$$-{\int }_0^{\frac{\pi }{2}}(-\mathrm{sinx})\frac{{\sin }^{\mathrm{n}+1}\mathrm{x}}{\mathrm{n}+1}\mathrm{dx}=\frac{1}{\mathrm{n}+1}{\int }_0^{\frac{\pi }{2}}{\sin }^{\mathrm{n}+2}\mathrm{xdx}$$$$=\frac{1}{\mathrm{n}+1}{\mathrm{I}}_{\mathrm{n}+2}\Rightarrow {\mathrm{I}}_{\mathrm{n}+2}={\mathrm{I}}_{\mathrm{n}}-\frac{1}{\mathrm{n}+1}{\mathrm{I}}_{\mathrm{n}+2}\Rightarrow$$$$(1+\frac{1}{\mathrm{n}+1}){\mathrm{I}}_{\mathrm{n}+2}={\mathrm{I}}_{\mathrm{n}}\Rightarrow \frac{\mathrm{n}+2}{\mathrm{n}+1}{\mathrm{I}}_{\mathrm{n}+2}={\mathrm{I}}_{\mathrm{n}}\Rightarrow {\mathrm{I}}_{\mathrm{n}+2}=\frac{\mathrm{n}+1}{\mathrm{n}+2}{\mathrm{I}}_{\mathrm{n}}$$$${\mathrm{I}}_0={\int }_0^{\frac{\pi }{2}}\mathrm{dx}=\frac{\pi }{2}$$$${\mathrm{I}}_1={\int }_0^{\frac{\pi }{2}}\mathrm{sinxdx}={[-\mathrm{cosx}]}_0^{\frac{\pi }{2}}=1$$

Answered by mathmax by abdo last updated on 20/Dec/21

$$\mathrm{W}={\int }_0^{\sqrt{2}-1}\frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}}}{\mathrm{x}+1}\mathrm{dx}\Rightarrow \mathrm{W}={\int }_0^{\sqrt{2}-1}\frac{\sqrt{{\mathrm{x}}^2+\mathrm{x}+\frac{1}{4}-\frac{1}{4}}}{\mathrm{x}+1}\mathrm{dx}$$$$={\int }_0^{\sqrt{2}-1}\frac{\sqrt{{(\mathrm{x}+\frac{1}{2})}^2-\frac{1}{4}}}{\mathrm{x}+1}\mathrm{dx}$$$${=}_{\mathrm{x}+\frac{1}{2}=\frac{\mathrm{chu}}{2}\to \mathrm{u}=\mathrm{argch}(2\mathrm{x}+1)=\ln (2\mathrm{x}+1+\sqrt{{(2\mathrm{x}+1)}^2-1})}{\int }_0^{\ln (2\sqrt{2}-1+\sqrt{{(2\sqrt{2}-1)}^2-1})}\frac{\mathrm{shu}}{2(\frac{\mathrm{chu}}{2}+1)}\frac{1}{2}\mathrm{shudu}$$$$=\frac{1}{2}{\int }_0^{\ln (2\sqrt{2}-1+\sqrt{{(2\sqrt{2}-1)}^2-1})}\frac{{\mathrm{sh}}^2\mathrm{u}}{2+\mathrm{chu}}\mathrm{du}\mathrm{let}\mathrm{find}$$$$\int \frac{{\mathrm{sh}}^2\mathrm{u}}{2+\mathrm{chu}}\mathrm{du}=\int \frac{\mathrm{ch}\left(2\mathrm{u}\right)-1}{2(2+\mathrm{chu})}\mathrm{du}=\int \frac{\frac{{\mathrm{e}}^{2\mathrm{u}}+{\mathrm{e}}^{-2\mathrm{u}}}{2}-1}{2(2+\frac{{\mathrm{e}}^{\mathrm{u}}+{\mathrm{e}}^{-\mathrm{u}}}{2})}\mathrm{du}$$$$=\frac{1}{2}\int \frac{{\mathrm{e}}^{2\mathrm{u}}+{\mathrm{e}}^{-2\mathrm{u}}-2}{4+{\mathrm{e}}^{\mathrm{u}}+{\mathrm{e}}^{-\mathrm{u}}}\mathrm{du}{=}_{{\mathrm{e}}^{\mathrm{u}}=\mathrm{x}}\frac{1}{2}\int \frac{{\mathrm{x}}^2+{\mathrm{x}}^{-2}-2}{4+\mathrm{x}+{\mathrm{x}}^{-1}}\frac{\mathrm{dx}}{\mathrm{x}}$$$$=\frac{1}{2}\int \frac{{\mathrm{x}}^2+{\mathrm{x}}^{-2}}{{\mathrm{x}}^2+4\mathrm{x}+1}\mathrm{dx}=\frac{1}{2}\int \frac{{\mathrm{x}}^4+1}{{\mathrm{x}}^2({\mathrm{x}}^2+4\mathrm{x}+1)}\mathrm{dx}$$$$\mathrm{rest}\mathrm{decomposition}....\mathrm{be}\mathrm{continued}...$$

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