A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{4}\]
Correct Answer: D
Solution :
Let \[I=\int_{0}^{\pi /2}{\frac{\sin x}{\sin x+\cos x}dx}\] ... (i) \[=\int_{0}^{\pi /2}{\frac{\sin \left( \frac{\pi }{2}-x \right)}{\sin \left( \frac{\pi }{2}-x \right)+\cos \left( \frac{\pi }{2}-x \right)}dx}\] \[\Rightarrow \] \[I=\int_{0}^{\pi /2}{\frac{\cos x}{\sin x+\cos x}dx}\] ? (ii) On adding Eqs. (i) and (ii), we get \[2I=\int_{0}^{\pi /2}{1dx}\] \[=[x]_{0}^{\pi /2}=\frac{\pi }{2}\] \[\Rightarrow \] \[I=\frac{\pi }{4}\]
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