A) \[\pi \]
B) \[\pi /2\]
C) \[4\pi \]
D) \[\pi /4\]
Correct Answer: D
Solution :
\[I=\int\limits_{\pi /2}^{\pi /2}{\frac{{{\sin }^{2}}x}{1+{{2}^{x}}}dx}\] ?(i) \[\Rightarrow \]\[I=\int\limits_{-\pi /2}^{\pi /2}{\frac{{{\sin }^{2}}x}{1+{{2}^{-x}}}dx,}\]by replacing x by \[\left( \frac{\pi }{2}-\frac{\pi }{2}-x \right)\]\[\Rightarrow \]\[I=\int\limits_{-\pi /2}^{\pi /2}{\frac{{{2}^{x}}.{{\sin }^{2}}x}{1+{{2}^{x}}}dx}\] ?(ii) Adding equations (i) and (ii), we get \[2I=\int\limits_{-\pi /2}^{\pi /2}{{{\sin }^{2}}xdx=\frac{1}{2}\int\limits_{-\pi /2}^{\pi /2}{(1-\cos 2x)dx}}\] \[\Rightarrow \]\[I=\frac{1}{4}\left[ x+\frac{\sin 2x}{2} \right]_{-\pi /2}^{\pi /2}\] \[=\frac{1}{4}\left[ \left( \frac{\pi }{2}+\frac{\sin \pi }{2} \right)-\left( -\frac{\pi }{2}+\frac{\sin (-\pi )}{2} \right) \right]\] \[I=\frac{1}{4}\left[ \frac{\pi }{2}+\frac{\pi }{2} \right]=\frac{\pi }{4}\]
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