$\int_{\pi/6}^{\pi/3} (\sin x + \cos x) / \sqrt{\sin(2x)}\, dx$ Question How to solve the following integral?
$$I = \int_{\pi/6}^{\pi/3} \dfrac{\sin x + \cos x}{\sqrt{\sin(2x)}}\,dx$$
Attempt Using the property of definite integrals and putting $\pi/6 + \pi/3 - x$ in place of $x$ did not help. So I tried simplifying the function and I got $[(\tan x)^{1/2} + (\cot x)^{1/2}]/2^{1/2}$, with which I don't know how to proceed?
Any helps appreciated. 

The interval $\left(\frac{\pi}{6},\frac{\pi}{3}\right)$ is centered at $\frac{\pi}{4}$ and $\sin(x)+\cos(x)= \sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$, hence the given integral can be written as
$$ \int_{-\pi/12}^{\pi/12}\frac{\sqrt{2}\sin\left(x+\frac{\pi}{2}\right)}{\sqrt{\sin\left(2x+\frac{\pi}{2}\right)}}\,dx=\sqrt{2}\int_{-\pi/12}^{\pi/12}\frac{\cos(x)}{\sqrt{\cos(2x)}}\,dx=2\sqrt{2}\int_{0}^{\pi/12}\frac{\cos(x)}{\sqrt{\cos(2x)}}\,dx$$
or as
$$ \sqrt{2}\int_{0}^{\pi/6}\frac{\cos(x/2)}{\sqrt{\cos x}}\,dx=\int_{0}^{\pi/6}\sqrt{\frac{1+\cos x}{\cos x}}\,dx=\int_{\sqrt{3}/2}^{1}\sqrt{\frac{1+x}{x}}\frac{dx}{\sqrt{1-x^2}} $$
or as
$$ \int_{\sqrt{3}/2}^{1}\frac{dx}{\sqrt{x(1-x)}}=2\int_{\sqrt{\sqrt{3}/2}}^{1}\frac{du}{\sqrt{1-u^2}}=\pi-2\arcsin\sqrt{\frac{\sqrt{3}}{2}}=\boxed{\arccos(\sqrt{3}-1)}. $$
