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Question Number 26507 by yesaditya22@gmail.com last updated on 26/Dec/17

$${\stackrel{\mathrm{\Pi }}{\int }}_0\frac{\mathrm{xsinx}}{1+{\cos }^2\mathrm{x}}\mathrm{dx}$$

Answered by kaivan.ahmadi last updated on 26/Dec/17

$$\mathrm{I}={\int }_0^{\pi }\frac{\mathrm{xsinx}}{2-{\sin }^2\mathrm{x}}\mathrm{dx}.\mathrm{set}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{2-{\mathrm{x}}^2}\Rightarrow$$$$\mathrm{f}\left(\mathrm{sinx}\right)=\mathrm{f}\left(\sin (\pi -\mathrm{x})\right)\Rightarrow$$$${\int }_0^{\pi }\frac{\mathrm{xsinx}}{2-{\sin }^2\mathrm{x}}\mathrm{dx}=\frac{\pi }{2}{\int }_0^{\pi }\mathrm{f}\left(\mathrm{sinx}\right)\mathrm{dx}=$$$$\frac{\pi }{2}{\int }_0^{\pi }\frac{\mathrm{sinx}}{1+{\cos }^2\mathrm{x}}\mathrm{now}\mathrm{if}\mathrm{u}=\mathrm{cosx}\Rightarrow \mathrm{du}=-\mathrm{sinxdx}$$$${\mathrm{I}=-\frac{\pi }{2}{\int }_1^{-1}\frac{\mathrm{du}}{1+{\mathrm{u}}^2}=-\frac{\pi }{2}\mathrm{Arctgu}]}_1^{-1}=$$$$-\frac{\pi }{2}(\mathrm{arctg}(-1)-\mathrm{arctg}1)=$$$$-\frac{\pi }{2}(-\frac{\pi }{4}-\frac{\pi }{4})=\frac{{\pi }^2}{4}$$$$\Rightarrow$$

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