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Question Number 106932 by Ar Brandon last updated on 07/Aug/20

$${\int }_0^{\pi }\frac{\mathrm{x}}{1+{\cos }^2\mathrm{x}}\mathrm{dx}$$

Answered by mathmax by abdo last updated on 08/Aug/20

$$\mathrm{I}={\int }_0^{\pi }\frac{\mathrm{xdx}}{1+{\cos }^2\mathrm{x}}\mathrm{changement}\mathrm{x}=\pi -\mathrm{t}\mathrm{give}$$$$\mathrm{I}={\int }_0^{\pi }\frac{(\pi -\mathrm{t})\mathrm{dt}}{1+{\cos }^2\mathrm{t}}=\pi {\int }_0^{\pi }\frac{\mathrm{dt}}{1+{\cos }^2\mathrm{t}}-\mathrm{I}\Rightarrow 2\mathrm{I}=\pi {\int }_0^{\pi }\frac{\mathrm{dt}}{1+{\cos }^2\mathrm{t}}$$$$\mathrm{but}\mathrm{we}\mathrm{have}\mathrm{proved}{\int }_0^{\pi }\frac{\mathrm{dt}}{1+{\cos }^2\mathrm{t}}=\frac{\pi }{\sqrt{2}}\Rightarrow 2\mathrm{I}=\frac{{\pi }^2}{\sqrt{2}}\Rightarrow \mathrm{I}=\frac{{\pi }^2}{2\sqrt{2}}$$

Answered by Dwaipayan Shikari last updated on 08/Aug/20

$${\int }_0^{\pi }\frac{x}{1+{cos}^{2}x}=\int \frac{\pi -x}{1+{cos}^{2}x}=I$$$$2I={\int }_0^{\pi }\frac{\pi }{1+{cos}^{2}x}$$$$2I=\pi {\int }_0^{\pi }\frac{{sec}^{2}x}{{sec}^{2}x+1}$$$$2I=\pi {\int }_0^{\pi }\frac{dt}{t^2+2}(tanx=t$$$$2I=\frac{\pi }{\sqrt{2}}{\left[{tan}^{-1}\left(\frac{\mathrm{tanx}}{\sqrt{2}}\right)\right]}_0^{\pi }=\frac{\pi }{\sqrt{2}}.\pi =\frac{{\pi }^2}{\sqrt{2}}$$$$I=\frac{{\pi }^2}{2\sqrt{2}}$$

Commented by Ar Brandon last updated on 08/Aug/20

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Answered by Ar Brandon last updated on 08/Aug/20

$$\mathcal{J}={\int }_0^{\pi }\frac{\mathrm{x}}{1+{\cos }^2\mathrm{x}}\mathrm{dx}$$$$={\int }_0^{\pi }\frac{\pi -\mathrm{x}}{1+{\cos }^2\mathrm{x}}\mathrm{dx}={\int }_0^{\pi }\frac{\pi }{1+{\cos }^2\mathrm{x}}\mathrm{dx}-\mathcal{J}$$$$2\mathcal{J}={\int }_0^{\pi }\frac{\pi }{1+{\cos }^2\mathrm{x}}\mathrm{dx}=\pi {\int }_0^{\pi }\frac{{\sec }^2\mathrm{x}}{{\sec }^2\mathrm{x}+1}\mathrm{dx}$$$$=\pi {\int }_0^{\pi }\frac{{\sec }^2\mathrm{x}}{{\tan }^2\mathrm{x}+2}\mathrm{dx}=\pi {\int }_0^{\pi }\frac{\mathrm{d}\left(\mathrm{tanx}\right)}{{\tan }^2\mathrm{x}+2}\mathrm{dx}$$$$=\pi {\left[\frac{1}{\sqrt{2}}\mathrm{Arctan}\left(\frac{\mathrm{tanx}}{\sqrt{2}}\right)\right]}_0^{\pi }$$$$=\frac{\pi }{\sqrt{2}}[\mathrm{Arctan}(-\frac{0}{\sqrt{2}})-\mathrm{Arctan}\left(\frac{0}{\sqrt{2}}\right)]=\frac{\pi (\pi -0)}{\sqrt{2}}=\frac{{\pi }^2}{\sqrt{2}}$$$$\mathcal{J}=\frac{{\pi }^2}{2\sqrt{2}}$$

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