∫limits_(π)/6^(π)/3 (d x)/(1+√(tan x)) is equal to :

$\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\tan x}}=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\frac{\sqrt{\sin x}}{\sqrt{\cos x}}} d x$

$I=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x$       ........(1)

Multiplying and dividing by $\sqrt{\cos x}$

$I=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos (\frac{\pi}{3}+\frac{\pi}{6}-x)} d x}{\sqrt{\cos (\frac{\pi}{3}+\frac{\pi}{6}-x)}+\sqrt{\sin (\frac{\pi}{3}+\frac{\pi}{6}-x)}}$

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

$I=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos (\frac{\pi}{2}-x)}}{\sqrt{\cos (\frac{\pi}{2}-x)}+\sqrt{\sin (\frac{\pi}{2}-x)}} d x$

$I=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\cos x} +\sqrt{\sin x}} d x$         .......(2)

addding (1) and (2)

$2 I=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}+\sqrt{\sin x} d x}{\sqrt{\cos x}+\sqrt{\sin x}}= \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx$

$2 I=[x]_{\frac{\pi}{6}}^{\frac{\pi}{3}}=\frac{\pi}{3}-\frac{\pi}{6}$

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

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