The value of $\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{1+\tan x}$ is:

$\frac{\pi}{12}$

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

$\int\limits_a^bf(x)dx=\int\limits_a^bf(a+b-x)dx$

$I=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\tan(\frac{\pi}{2}-x)}dx$

$I=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\cot x}dx=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\tan x}dx$

$∴2I=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left(\frac{1}{1+\tan x}+\frac{1}{1+\cot x}\right)dx$

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

$2I=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}}1dx=\left(\frac{\pi}{3}-\frac{\pi}{6}\right)=\frac{\pi}{6}$

$∴I=\frac{\pi}{12}$