A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{12}\]
C) \[\frac{2\pi }{3a}\]
D) \[\frac{\pi }{4}\]
Correct Answer: B
Solution :
[b] \[I=\int\limits_{\pi /6}^{\pi /3}{\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}.dx}\] \[=\int\limits_{\pi /6}^{\pi /3}{\frac{\sqrt{\sin \left( \frac{\pi }{3}+\frac{\pi }{6}-x \right)}}{\sqrt{\sin \left( \frac{\pi }{3}+\frac{\pi }{6}-x \right)}+\sqrt{\cos \left( \frac{\pi }{3}+\frac{\pi }{6}-x \right)}}.dx}\] \[=\int\limits_{\pi /6}^{\pi /3}{\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}.dx}\] \[2I=\int\limits_{\pi /6}^{\pi /3}{.dx}=[x]_{\pi /6}^{\pi /3}=\left[ \frac{\pi }{3}-\frac{\pi }{6} \right]=\frac{\pi }{6}\] \[\because I=\frac{\pi }{12}\] Hence, option [b] is correct.
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