Question

$\underset{\mathrm{\pi }/6}{\overset{\mathrm{\pi }/3}{\int }}\frac{1}{1+\sqrt{\tan x}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{1}{1+\sqrt{\tan x}}dx...\left(\mathrm{i}\right) \\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{1}{1+\sqrt{\tan \left(\frac{\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6}-x\right)}}dx \\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{1}{1+\sqrt{cotx}}dx...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right) \\ 2I={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\left(\frac{1}{1+\sqrt{\tan x}}+\frac{1}{1+\sqrt{cotx}}\right)dx \\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\left(\frac{1+\sqrt{cotx}+1+\sqrt{\tan x}}{1+\sqrt{cotx}+\sqrt{\tan x}+\sqrt{\tan xcotx}}\right)dx \\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{2+\sqrt{cotx}+\sqrt{\tan x}}{2+\sqrt{cotx}+\sqrt{\tan x}}dx \\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}dx={\left[x\right]}_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}} \\ =\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{6} \\ \therefore 2I=\frac{\mathrm{\pi }}{6} \\ \mathrm{Hence}I=\frac{\mathrm{\pi }}{12} \end{gathered}$$