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.
 A: 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, $$I=\color{red}{\frac{\pi}{12}}.$$
A: Heh. Sometimes there are ways to find a definite integral without finding an antiderivative.
Note that $\tan(\pi/2-t)=1/\tan(t)$. Say $I$ is your integral. The change of variables $t\mapsto\pi/2-t$ shows that $$I=\int_{\pi/6}^{\pi/3}\frac{1}{1+\sqrt{1/\tan(t)}}\,dt=\int_{\pi/6}^{\pi/3}\frac{\sqrt{\tan(t)}}{1+\sqrt{\tan(t)}}\,dt.$$
Hence $$I+I=\int_{\pi/6}^{\pi/3}\,dt.$$(That's as far as I got, sorry...)
A: $$\begin{align}\int_{\pi/6}^{\pi/3}\frac{dx}{1 + \sqrt{\tan x}} &= \int_{\pi/6}^{\pi/4}\frac{dx}{1 + \sqrt{\tan x}} +\int_{\pi/4}^{\pi/3}\frac{dx}{1 + \sqrt{\tan x}}\\
&=\int_{\pi/6}^{\pi/4}\frac{dx}{1 + \sqrt{\tan x}} +\int_{\pi/6}^{\pi/4}\frac{\sqrt {\tan t}\, dt}{1 + \sqrt{\tan t}}, \quad t = \pi/2 -x\\
&=\int_{\pi/6}^{\pi/4} 1 \, dx\\
 &= \frac{\pi}{12}
\end{align}$$
