The value of the integral ${\int }_{\pi /6}^{\pi /3} \frac{1}{1+\sqrt{\tan x}}dx$, is

Let $I={\int }_{\pi /6}^{\pi /3} \frac{1}{1+\sqrt{\tan x}}dx$                                    …(i)

$\Rightarrow l={\int }_{\pi /6}^{\pi /3} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx$                       ..(ii)

This is of the form ${\int }_a^b \frac{f\left(x\right)}{f\left(x\right)+f(a+b-x)}dx$

$\therefore I=\frac{\pi /3-\pi /6}{2}=\frac{\pi }{12}$.