A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{2}\]
C) \[\pi \]
D) \[2\pi \]
Correct Answer: A
Solution :
\[I=\int_{\,0}^{\,\pi /2}{\frac{{{2}^{\sin x}}}{{{2}^{\sin x}}+{{2}^{\cos x}}}dx}\] ?..(i) \[I=\int_{0}^{\,\pi /2}{\frac{{{2}^{\sin \left( \frac{\pi }{2}-x \right)}}}{{{2}^{\sin \left( \frac{\pi }{2}-x \right)}}+{{2}^{\cos \left( \frac{\pi }{2}-x \right)}}}dx}\]\[=\int_{0}^{\pi /2}{\frac{{{2}^{\cos x}}}{{{2}^{\cos x}}+{{2}^{\sin x}}}}\,dx\] ?..(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, \[I=\frac{\pi }{4}\].
You need to login to perform this action.
You will be redirected in
3 sec
