Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\mathrm{tanx}}}dx...\left(\mathrm{i}\right) \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\cot \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}}{\sqrt{\cot \left(\left(\frac{\mathrm{\pi }}{2}-x\right)\right)}+\sqrt{\tan \left(\frac{\mathrm{\pi }}{2}-x\right)}}dx\left[\mathrm{Using}{\int }_0^{\mathrm{a}}\mathrm{f}\left(x\right)\mathrm{dx}={\int }_0^{\mathrm{a}}\mathrm{f}\left(a\mathit{-}x\right)dx\right] \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\mathrm{tanx}}}{\sqrt{\mathrm{tanx}}+\sqrt{\mathrm{cotx}}}\mathit{}dx\mathit{}\mathit{}...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right) \\ 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\frac{\sqrt{\mathrm{cotx}}}{\sqrt{\mathrm{cotx}}+\sqrt{\mathrm{tanx}}}+\frac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}}\right)\mathit{d}x\mathit{}\mathit{} \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}dx \\ ={\left[x\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ =\frac{\mathrm{\pi }}{2} \\ \mathrm{Hence},I=\frac{\mathrm{\pi }}{4} \end{gathered}$$​