Definite integral $\int_0^{\frac{\pi}{2}}\! \frac{\sin x }{\sqrt{\sin 2x}}\,\mathrm{d} x$ $$\int_0^{\frac{\pi}{2}}\! \frac{\sin x }{\sqrt{\sin 2x}}\,\mathrm{d} x$$
I'm pretty sure I can finish it after finding the anti derivative. I tried changing the denominator to $2\sin x \cos x$ and subbing $u$ as $\sin x$.
$$\int_0^{\frac{\pi}{2}}\! \frac{\sin x }{\sqrt{2\sin x\cos x}}\,\mathrm{d} x$$
 A: This can be written as
$$\dfrac{1}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \sqrt{\tan x}\, dx$$
Now substitute $t^2 = \tan x$ $-$ this is valid since $\tan x \ge 0$ for $0 \le x < \frac{\pi}{2}$.
Then $2t\, dt = (1+t^4)\, dx$, and so the integral is equal to
$$\dfrac{1}{\sqrt{2}} \int_0^{\infty} t \cdot \dfrac{2t}{1+t^4}\, dt$$
Try to take over from here; let me know if you have more problems.
You should end up with $\dfrac{\pi}{2}$, and it is possible to find a closed form for the antiderivative.
A: After a little algebra, we get 
$$ \int_0^{ \frac{\pi}{2}}\! dx \: \frac{\sin x }{\sqrt{\sin 2x}} = \frac{1}{\sqrt{2}} \int_0^{\frac{\pi}{2}} dx \: \frac{\sin{x}}{\sqrt{\cos{x} \sqrt{1-\cos^2{x}}}}$$
Substitute $u=\cos{x}$:
$$\int_0^{ \frac{\pi}{2}}\! dx \: \frac{\sin x }{\sqrt{\sin 2x}} = \frac{1}{\sqrt{2}} \int_0^1 du \: u^{-1/2} (1-u^2)^{-1/4} $$
Now substitute $v = u^2$:
$$\begin{align} \int_0^{ \frac{\pi}{2}}\! dx \: \frac{\sin x }{\sqrt{\sin 2x}} &= \frac{1}{2 \sqrt{2}} \int_0^1 dv \: v^{-3/4} (1-v)^{-1/4} \\ &= \frac{1}{2 \sqrt{2}} \Gamma \left( \frac{1}{4} \right ) \Gamma \left( \frac{3}{4} \right ) \\ &= \frac{\pi}{2} \end{align} $$
That last step comes from the relation
$$\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin{\pi z}}$$
