Finding $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\sqrt{\tan x}}dx$ 
How can I integrate $$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\sqrt{\tan x}}dx\ \ \ ?$$

I have made the integral into the form of $\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}$, but been able to go no further.

Let $I$ be your integral. 
Setting $u=\frac{\pi}{2}-x$ and using $\tan\left(\frac{\pi}{2}-x\right)=\frac{1}{\tan x}$, we have
$$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\sqrt{\tan x}}dx=\int_{\frac{\pi}{3}}^{\frac{\pi}{6}}\frac{-1}{1+\sqrt{\frac{1}{\tan u}}}du=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\sqrt{\tan u}}{1+\sqrt{\tan u}}du=J.$$
Now note that 
$$I+J=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}.$$
Hence, $$I=\color{red}{\frac{\pi}{12}}.$$
