$\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{cot\, x}}dx$ is :
Answer & explanation
Correct answer: option 3
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)
$2I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1+\sqrt{\cot x}}{1+\sqrt{\cot x}}dx=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}1dx$
$2I=\left[\frac{\pi}{3}-\frac{\pi}{6}\right]$
$2I=\frac{\pi}{6}$
$I=\frac{\pi}{12}$