A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{3}\]
Correct Answer: C
Solution :
\[I=\int_{0}^{\pi /2}{\frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}}dx}\] .....(i) \[=\int_{0}^{\pi /2}{\frac{\sqrt{\cot \left( \frac{\pi }{2}-x \right)}}{\sqrt{\cot \left( \frac{\pi }{2}-x \right)}+\sqrt{\tan \left( \frac{\pi }{2}-x \right)}}dx}\] \[=\int_{0}^{\pi /2}{\frac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}}dx}\] .....(ii) Now adding (i) and (ii), we get \[2I=\int_{0}^{\pi /2}{\frac{\sqrt{\cot x}+\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}}dx=[x]_{0}^{\pi /2}\Rightarrow I=\frac{\pi }{4}}\].
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