how to evaluate $\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin x}}\text{d}x$ I was solving a physics problem and eventually the problem boiled down to solving the following integral:

$$\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin x}}\text{d}x$$

I have already tried substitutions like $\sin x=t^2$ , $\sin x=t$ and have tried using the properties of definite integrals given on http://www.sosmath.com/calculus/integ/integ02/integ02.html but I could not solve this integral. Please help!
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$\ds{{\cal I} \equiv \int_{0}^{\pi/2}{\dd x \over \root{\sin\pars{x}}}:\ {\large ?}}$


With $x = \arcsin\pars{z^{2}}$ we'll get:
$$
{\cal I} = 2\int_{0}^{1}{\dd z \over \root{1 - z^{4}}}\,.\quad\mbox{With}\ z = t^{1/4}\,,\quad  
{\cal I} = \half\int_{0}^{1}t^{-3/4}\pars{1 - t}^{-1/2}\,\dd t
$$


Then
$$
{\cal I}=\half\,{\rm B}\pars{{1 \over 4},\half}=\half\,{\Gamma\pars{1/4}\Gamma\pars{1/2} \over \Gamma\pars{3/4}}
={\root{\pi} \over 2}\,\Gamma^{2}\pars{1/4}\,{\sin\pars{\pi/4} \over \pi} 
$$
$$\color{#00f}{\large%
\int_{0}^{\pi/2}{\dd x \over \root{\sin\pars{x}}}
=
{1 \over 4}\,\root{2 \over \pi}\,\Gamma^{\,2}\pars{1 \over 4}}
\approx 2.62206
$$
So, which answer is prefereable?
$$
\sqrt{2}\;\mathrm{K}\left(\frac{1}{\sqrt{2}}\right) = 
\frac{1}{2}\;\mathrm{B}\left(\frac{1}{4},\frac{1}{2}\right) =
\frac{\pi^{3/4}}{\sqrt{2}\,\Gamma(3/4)^2}
$$