A) \[\frac{1}{2}\log 2\]
B) \[\frac{\pi }{2}-\log 2\]
C) \[\frac{\pi }{4}-\frac{1}{2}\log 2\]
D) \[\frac{\pi }{2}+\log 2\]
Correct Answer: C
Solution :
[c] \[I=\int\limits_{0}^{\pi /2}{\frac{\sin xdx}{1+\sin x+\cos x}}\] \[=\int\limits_{0}^{\pi /2}{\frac{\cos xdx}{1+\sin x+\cos x}}\] Or \[2I=\int\limits_{0}^{\pi /2}{\frac{\sin x+\cos x+1-1}{\sin x+\cos x+1}dx}\] \[2I=\frac{\pi }{2}-\log 2\] Or \[I=\frac{\pi }{4}-\frac{1}{2}\log 2\]
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