A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{6}\]
Correct Answer: C
Solution :
\[I=\int_{0}^{\pi /2}{\frac{\sin x.dx}{\sin x+\cos x}}=\int_{0}^{\pi /2}{\frac{\cos x.dx}{\cos x+\sin x}}\], \[\,\,\left( \because \int_{0}^{a}{f(x)dx=\int_{0}^{a}{f(a-x)dx}} \right)\] \[2I=\int_{0}^{\pi /2}{dx}\Rightarrow I=\frac{\pi }{4}\].
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