$\int\limits^{\frac{\pi }{3}}_{\frac{\pi }{6}}\frac{1}{1+\sqrt{cotx}}dx$ is equal to :

$\frac{\pi }{12}$

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

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

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

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

adding (1) and (2)

$2I=\int\limits^{\frac{\pi }{3}}_{\frac{\pi }{6}}1dx⇒I=\frac{\pi }{12}$