Question

\[\int\limits_0^{\pi/2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} dx\]

Answer 1

\[Let\ I = \int_0^\frac{\pi}{2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{tanx}} d x ................(1) \]

\[ = \int_0^\frac{\pi}{2} \frac{\sqrt{\cot\left( \frac{\pi}{2} - x \right)}}{\sqrt{\cot\left( \left( \frac{\pi}{2} - x \right) \right)} + \sqrt{\tan\left( \frac{\pi}{2} - x \right)}} dx ................\left[\text{Using }\int_0^a f\left( x \right) dx = \int_0^a f\left( a - x \right) dx \right]\]

\[ = \int_0^\frac{\pi}{2} \frac{\sqrt{tanx}}{\sqrt{tanx} + \sqrt{cotx}} dx ..............(2)\]

\[ \text{Adding (1) and (2})\]

\[2I = \int_0^\frac{\pi}{2} \left( \frac{\sqrt{cotx}}{\sqrt{cotx} + \sqrt{tan x}} + \frac{\sqrt{\tan x}}{\sqrt{\tan x} + \sqrt{\cot x}} \right) d x \]

\[ = \int_0^\frac{\pi}{2} dx \]

\[ = \left[ x \right]_0^\frac{\pi}{2} \]

\[ = \frac{\pi}{2}\]

\[Hence\, I = \frac{\pi}{4}\]