Given $I=\int \limits _{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt {\tan x}}$-----(1)
This can be written as
$I=\int \limits _{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt {\sin x}}=\int \limits _{\pi/6}^{\pi/3} \frac{\sqrt \cos x}{\sqrt {\cos x}+\sqrt {\sin x}}dx$
Applying the property $ \int \limits_a^b f(x)dx=\int \limits_a^b f(a+b-x) dx$
$I=\int \limits _{\pi/6}^{\pi/3} \large\frac{\sqrt {\cos (\pi/6+\pi/3-x)}}{\sqrt {\cos (\pi/6+\pi/3-x)}+\sin \sqrt {(\pi/6+\pi/3-x)}}dx$
$I=\int \limits _{\pi/6}^{\pi/3} \frac{\sqrt {\cos (\pi/2-x)}}{\sqrt {\cos (\pi/2-x)}+\sqrt {(\pi/2-x)}}dx$
But $\cos (\pi/2-x)=\sin x$ and $ \sin (\pi/2-x)=\cos x$
Therefore $I=\int \limits _{\pi/6}^{\pi/3} \frac{\sqrt {\sin x}}{\sqrt {\sin x}+\sqrt {\cos x}}dx$-----(2)
Adding equ(1) and equ (2)
$I=\int \limits _{\pi/6}^{\pi/3} \frac{\sqrt {\sin x}+\sqrt {\cos x }}{\sqrt {\sin x}+\sqrt {\cos x}}dx=\int \limits _{\pi/6}^{\pi/3} dx $
on integrating we get,
$2I=\int \limits _{\pi/6}^{\pi/3} dx$
$=[x]_{\pi/6}^{\pi/3} $
Applying the limits we get
$2I=\bigg[\frac{\pi}{3}-\frac{\pi}{6}\bigg]$
$=\frac{\pi}{6}$
Therefore $I=\frac{\pi}{12}$
