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
$$
{\cal I}=\half\,{\rm B}\pars{{1 \over 4},\half}=\half\,{\Gamma\pars{1/4}\Gamma\pars{1/2} \over \Gamma\pars{3/4}}
={\root{\pi} \over 2}\,\Gamma^{2}\pars{1/4}\,{\sin\pars{\pi/4} \over \pi} 
$$
$$\color{#00f}{\large%
\int_{0}^{\pi/2}{\dd x \over \root{\sin\pars{x}}}
=
{1 \over 4}\,\root{2 \over \pi}\,\Gamma^{\,2}\pars{1 \over 4}}
\approx 2.62206
$$
A: Use the change $\sin(x)=\sqrt t$:
$$I=\displaystyle{1\over 2}\int_0^1 {t^{-{3\over 4}}\over\sqrt{1-t}}dt =
{1\over 2}\beta\left({1\over 4},{1\over 2}\right).$$
(change guessed after calculating the integral with Maxima)
A: Let $u=\sin x$, then $t=u^2$, and recognize the expression of the beta function in the new integral.
A: $$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \frac{d x}{\sqrt{\sin x}} & =\int_0^{\frac{\pi}{2}} \sin ^{2\left(\frac{1}{4}\right)-1 }\theta \cos^{ 2\left(\frac{1}{2}\right)-1 }\theta d\theta\\
& =\frac{1}{2} B\left(\frac{1}{4}, \frac{1}{2}\right) \\
& =\frac{1}{2} \cdot \frac{\Gamma\left(\frac{1}{4}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{3}{4}\right)} \\
& =\frac{\sqrt{\pi}}{2} \cdot \frac{\Gamma\left(\frac{1}{4}\right) \cdot \Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right) \cdot \Gamma\left(\frac{1}{4}\right)} \\
& =\frac{\sqrt{\pi}}{2} \cdot \frac{\Gamma^2\left(\frac{1}{4}\right)}{\pi \csc \left(\frac{\pi}{4}\right)} \\
& =\frac{1}{2 \sqrt{2 \pi}} \Gamma^2\left(\frac{1}{4}\right)
\end{aligned}
$$
where the last step comes from the reflection property of Gamma function.
A: So, which answer is prefereable?
$$
\sqrt{2}\;\mathrm{K}\left(\frac{1}{\sqrt{2}}\right) = 
\frac{1}{2}\;\mathrm{B}\left(\frac{1}{4},\frac{1}{2}\right) =
\frac{\pi^{3/4}}{\sqrt{2}\,\Gamma(3/4)^2}
$$
