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 $\sin x$.

$$\int_0^{\frac{\pi}{2}}\! \frac{\sin x }{\sqrt{2\sin x\cos x}}\,\mathrm{d} x$$

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$\sin(x)=(\sqrt{\sin(x)})^2$ and then ask W|A – draks ... Jan 29 '13 at 23:11

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.

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$\displaystyle{\large{ {t^2 \over 1 + t^4} = \Re\left({1 \over t^2 + {\rm i}}\right) }}$. – Felix Marin Jul 11 '14 at 19:12

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

$$\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin{\pi z}}$$

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