$\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{1+\sqrt{\tan x}}$ is equal to

$\frac{\pi}{12}$

The correct answer is Option (4) → $\frac{\pi}{12}$

Let

$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{\tan x}}$

Use substitution $x=\frac{\pi}{2}-t$.

Then $\tan x=\cot t=\frac{1}{\tan t}$ and limits interchange.

$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dt}{1+\sqrt{\cot t}}$

$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan t}}{\sqrt{\tan t}+\;1}dt$

Add both forms:

$2I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left(\frac{1}{1+\sqrt{\tan x}}+\frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}\right)dx$

$2I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1\,dx$

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

$I=\frac{\pi}{12}$

Final answer: $\frac{\pi}{12}$