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Question Number 113738 by bemath last updated on 15/Sep/20

$$\underset{0}{\stackrel{\pi }{\int }}\frac{x\sin x}{1+\cos {}^{2}x}dx?$$

Answered by bobhans last updated on 15/Sep/20

$$\mathrm{I}=\underset{0}{\stackrel{\pi }{\int }}\frac{\mathrm{x}\sin \mathrm{x}}{1+\cos {}^2\mathrm{x}}\mathrm{dx}$$$$\mathrm{replace}\mathrm{x}\mathrm{by}\pi -\mathrm{x}\to \mathrm{I}=\underset{\pi }{\stackrel{0}{\int }}\frac{(\pi -\mathrm{x})\sin (\pi -\mathrm{x})}{1+\cos {}^2(\pi -\mathrm{x})}(-\mathrm{dx})$$$$\mathrm{I}=\underset{0}{\stackrel{\pi }{\int }}\frac{(\pi -\mathrm{x})\sin \mathrm{x}}{1+\cos {}^2\mathrm{x}}\mathrm{dx}=\underset{0}{\stackrel{\pi }{\int }}\frac{\pi \sin \mathrm{x}}{1+\cos {}^2\mathrm{x}}\mathrm{dx}-\underset{0}{\stackrel{\pi }{\int }}\frac{\mathrm{xsin}\mathrm{x}}{1+\cos {}^2\mathrm{x}}\mathrm{dx}$$$$2\mathrm{I}=\underset{0}{\stackrel{\pi }{\int }}\frac{\pi \sin \mathrm{x}}{1+\cos {}^2\mathrm{x}}\mathrm{dx}$$$$\mathrm{consider}\int \frac{\pi \sin \mathrm{x}}{1+\cos \mathrm{x}}\mathrm{dx}=-\pi \int \frac{\mathrm{d}\left(\cos \mathrm{x}\right)}{1+\cos {}^2\mathrm{x}}$$$$=-\pi \int \frac{\mathrm{du}}{1+{\mathrm{u}}^2}$$$$=-\pi {\tan }^{-1}\left(\cos \mathrm{x}\right)+\mathrm{c}$$$$\mathrm{now}\mathrm{we}\mathrm{have}2\mathrm{I}=-\pi {\left[{\tan }^{-1}\left(\cos \mathrm{x}\right)\right]}_0^{\pi }$$$$\mathrm{I}=-\frac{\pi }{2}[-\frac{\pi }{4}-\frac{\pi }{4}]=\frac{{\pi }^2}{4}.$$$$$$

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