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Question Number 146791 by puissant last updated on 15/Jul/21

$$\int \frac{\mathrm{x}}{1+{\cos }^2\left(\mathrm{x}\right)}\mathrm{dx}$$

Answered by Ar Brandon last updated on 15/Jul/21

$$\mathrm{I}=\int \frac{\mathrm{x}}{1+{\cos }^2\mathrm{x}}\mathrm{dx}=\int \frac{{\mathrm{xsec}}^2\mathrm{x}}{{\sec }^2\mathrm{x}+1}\mathrm{dx}$$$$=\int \frac{\mathrm{xd}\left(\mathrm{tanx}\right)}{2+{\tan }^2\mathrm{x}}=\frac{\mathrm{x}}{\sqrt{2}}\arctan \left(\frac{\mathrm{tanx}}{\sqrt{2}}\right)-\frac{1}{\sqrt{2}}\int \arctan \left(\frac{\mathrm{tanx}}{\sqrt{2}}\right)\mathrm{dx}$$$$\mathcal{I}={\int }_0^{\pi }\frac{\mathrm{x}}{1+{\cos }^2\mathrm{x}}\mathrm{dx}={\int }_0^{\pi }\frac{\pi -\mathrm{x}}{1+{\cos }^2\mathrm{x}}\mathrm{dx}$$$$=\pi {\int }_0^{\pi }\frac{\mathrm{dx}}{1+{\cos }^2\mathrm{x}}-\mathcal{I}=\pi {\int }_0^{\pi }\frac{\mathrm{d}\left(\mathrm{tanx}\right)}{2+{\tan }^2\mathrm{x}}-\mathcal{I}$$$$=\frac{\pi }{2}{\left[\frac{1}{\sqrt{2}}\arctan \left(\frac{\mathrm{tanx}}{\sqrt{2}}\right)\right]}_0^{\pi }=\frac{{\pi }^2\sqrt{2}}{4}$$

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