MCQ
The value of $\int_{\,0}^{\,\pi /2} {\frac{{{2^{\sin x}}}}{{{2^{\sin x}} + {2^{\cos x}}}}dx} $ is
- ✓$\frac{\pi }{4}$
- B$\frac{\pi }{2}$
- C$\pi $
- D$2\pi $
$I = \int_0^{\,\pi /2} {\frac{{{2^{\sin \left( {\frac{\pi }{2} - x} \right)}}}}{{{2^{\sin \left( {\frac{\pi }{2} - x} \right)}} + {2^{\cos \left( {\frac{\pi }{2} - x} \right)}}}}dx} $
$ = \int_0^{\pi /2} {\frac{{{2^{\cos x}}}}{{{2^{\cos x}} + {2^{\sin x}}}}} \,dx$....$(ii)$
Adding equations $(i)$ and $(ii),$ we get
$2I = \int_0^{\pi /2} {\left( {\frac{{{2^{\sin x}} + {2^{\cos x}}}}{{{2^{\sin x}} + {2^{\cos x}}}}} \right)dx = \int_{\,0}^{\,\pi /2} {1\,dx} = [x]\,_0^{\pi /2} = \frac{\pi }{2}} $
Therefore, $I = \frac{\pi }{4}$.
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