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

$\frac{\pi}{12}$

Let $I=\int\limits_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\tan x}} d x$          ...(i)

$\Rightarrow I=\int\limits_{\pi / 6}^{\pi / 3} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x$          ...(ii)

This is of the form $\int\limits_a^b \frac{f(x)}{f(x)+f(a+b-x)} d x$

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