$\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{cot\, x}}dx$ is :

$\frac{\pi }{12}$

The correct answer is Option (3) → $\frac{\pi }{12}$

$\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\cot x}}dx$  ...(1)

$I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\cot(\frac{\pi}{3}+\frac{\pi}{6}-x)}}dx$

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

$I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\cot x}}{\sqrt{\cot x}+1}dx$   ...(2)

Eq. (1) + Eq. (2)

$2I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1+\sqrt{\cot x}}{1+\sqrt{\cot x}}dx=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}1dx$

$2I=\left[\frac{\pi}{3}-\frac{\pi}{6}\right]$

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

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