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