A) \[\frac{\pi }{4}\]
B) \[2\pi \]
C) \[{{\pi }^{2}}\]
D) \[\frac{1}{2}{{\pi }^{2}}\]
Correct Answer: A
Solution :
\[I=\int_{0}^{\pi /2}{\frac{{{\sin }^{2}}x}{{{\sin }^{2}}x+{{\cos }^{2}}x}dx}\] ?..(i) \[\therefore \] \[I=\int_{0}^{\pi /2}{\frac{{{\sin }^{2}}\left( \frac{\pi }{2}-x \right)}{{{\sin }^{2}}\left( \frac{\pi }{2}-x \right)+{{\cos }^{2}}\left( \frac{\pi }{2}-x \right)}}dx\] \[\Rightarrow \] \[I=\int_{0}^{\pi /2}{\frac{{{\cos }^{2}}x}{{{\sin }^{2}}x+{{\cos }^{2}}x}}dx\] ?. (ii) On adding Eqs. (i) and (ii), we get \[2I=\int_{0}^{\pi /2}{1\,\,dx}\] \[\Rightarrow \] \[2I=[x]_{0}^{\pi /2}\] \[\Rightarrow \] \[I=\frac{\pi }{4}\]
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