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Question Number 18307 by mondodotto@gmail.com last updated on 18/Jul/17

Answered by ajfour last updated on 18/Jul/17

$$\mathrm{Q}.2$$$$\mathrm{I}={\int }_0^{\pi }\frac{\mathrm{ysin}\mathrm{ydy}}{1+\cos {}^2\mathrm{y}}={\int }_0^{\pi }\frac{(\pi -\mathrm{y})\sin \mathrm{ydy}}{1+\cos {}^2\mathrm{y}}$$$$=\pi {\int }_0^{\pi }\frac{\sin \mathrm{ydy}}{1+\cos {}^2\mathrm{y}}-\mathrm{I}$$$$\Rightarrow 2\mathrm{I}=-\pi {\tan }^{-1}\left(\cos \mathrm{y}\right){\mid }_0^{\pi }$$$$\mathrm{or}\mathrm{I}=-\frac{\pi }{2}[{\tan }^{-1}(-1)-{\tan }^{-1}1]$$$$=-\frac{\pi }{2}(-\frac{\pi }{4}-\frac{\pi }{4})$$$$\mathrm{I}=\frac{{\pi }^2}{4}.$$

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