Definite Integration Question 36
Question: The value of $ \int _{0}^{\pi /2}{\frac{{2^{\sin x}}}{{2^{\sin x}}+{2^{\cos x}}}dx} $ is
[Karnataka CET 1999; Kerala (Engg.) 2005]
Options:
A) $ \frac{\pi }{4} $
B) $ \frac{\pi }{2} $
C) $ \pi $
D) $ 2\pi $
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Answer:
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 ( \frac{\pi }{2}-x )}}}{{2^{\sin ( \frac{\pi }{2}-x )}}+{2^{\cos ( \frac{\pi }{2}-x )}}}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}{( \frac{{2^{\sin x}}+{2^{\cos x}}}{{2^{\sin x}}+{2^{\cos x}}} )dx=\int _{0}^{\pi /2}{1dx}=[x] _0^{\pi /2}=\frac{\pi }{2}} $
Therefore, $ I=\frac{\pi }{4} $ .
BETA
