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Question Number 120688 by bobhans last updated on 01/Nov/20

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

Answered by john santu last updated on 02/Nov/20

$$\psi =\underset{0}{\stackrel{\pi }{\int }}\frac{xdx}{1+\sin {}^{2}x};[letx=\pi -w]$$$$\psi =\underset{\pi }{\stackrel{0}{\int }}\frac{(\pi -x)(-dw)}{1+\sin {}^2{(\pi -w)}^2}=\underset{0}{\stackrel{\pi }{\int }}\frac{\pi -w}{1+\sin {}^{2}w}dw$$$$addingtheequationgive$$$$2\psi =\underset{0}{\stackrel{\pi }{\int }}\frac{\pi }{1+\sin {}^{2}x}dx$$$$\psi =\frac{1}{2}\underset{0}{\stackrel{\pi }{\int }}\frac{\pi }{1+\sin {}^{2}x}dx=\frac{\pi }{2}\underset{0}{\stackrel{\pi }{\int }}\frac{\mathrm{cosec}{}^{2}xdx}{\mathrm{cosec}{}^{2}x+1}$$$$\psi =\frac{\pi }{2}\underset{0}{\stackrel{\pi }{\int }}\frac{\mathrm{cosec}{}^{2}xdx}{(1+\cot {}^{2}x)+1}=\frac{\pi }{2}\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}\frac{1dt}{2+t^2};wheret=\cot x$$$$\psi =\frac{\pi }{2}.\frac{1}{\sqrt{2}}\mathrm{arc}\tan \left(\frac{t}{\sqrt{2}}\right){\mid }_{-\mathrm{\infty }}^{\mathrm{\infty }}$$$$\psi =\frac{\pi }{2\sqrt{2}}[\frac{\pi }{2}-(-\frac{\pi }{2})]=\frac{{\pi }^2}{2\sqrt{2}}.▴$$

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