Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{1}{1+\sqrt{\tan x}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+\sqrt{\tan x}}dx...\left(\mathrm{i}\right) \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+\sqrt{\tan \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}}dx\mathrm{Using}{\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+\sqrt{cotx}}dx...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\mathrm{we}\mathrm{get} \\ 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+\sqrt{\tan \mathrm{x}}}+\frac{1}{1+\sqrt{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}}d\mathrm{x} \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1+\sqrt{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}+1+\sqrt{\tan \mathrm{x}}}{\left(1+\sqrt{\tan \mathrm{x}}\right)\left(1+\sqrt{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}\right)}\mathrm{dx} \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1+\sqrt{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}+1+\sqrt{\tan \mathrm{x}}}{1+\sqrt{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}+\sqrt{\tan \mathrm{x}}+\sqrt{\mathrm{tanx}\mathrm{cotx}}}\mathrm{dx} \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{2+\sqrt{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}+\sqrt{\tan \mathrm{x}}}{2+\sqrt{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}+\sqrt{\tan \mathrm{x}}}\mathrm{dx} \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\mathrm{dx} \\ ={\left[\mathrm{x}\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ =\frac{\mathrm{\pi }}{2} \\ \mathrm{Hence}\mathit{}I=\frac{\mathrm{\pi }}{4} \end{gathered}$$