A) \[\frac{\pi }{12}\]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{6}\]
D) \[\frac{\pi }{4}\]
E) \[\frac{2\pi }{3}\]
Correct Answer: A
Solution :
Let \[\int_{\pi /6}^{\pi /3}{\frac{dx}{1+\sqrt{\tan x}}}\] \[\Rightarrow \] \[I=\int_{\pi /6}^{\pi /3}{\left( \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} \right)d}x\] ?.(i) \[=\int_{\pi /6}^{\pi /3}{\frac{\sqrt{\cos \left( \frac{\pi }{2}-x \right)}}{\sqrt{\sin \left( \frac{\pi }{2}-x \right)+\sqrt{\cos \left( \frac{\pi }{2}-x \right)}}}}dx\] \[=\int_{\pi /6}^{\pi /3}{\frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}}}dx\] ?.. (ii) On adding Eqs. (i) and (ii), we get \[2I=\int_{\pi /6}^{\pi /3}{1}dx=[x]_{\pi /6}^{\pi /3}=\frac{\pi }{3}-\frac{\pi }{6}=\frac{\pi }{6}\] \[\Rightarrow \] \[I=\frac{\pi }{12}\]
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