Compute the integral $\int_0^{\frac\pi2}\frac{\mathrm d \theta}{\sqrt{\sin \theta}}$ Compute the Riemann integral
$$\int_0^{\frac\pi2}\frac{\mathrm d \theta}{\sqrt{\sin \theta}}$$
It seems very difficult, I don't know how to go ahead.
Thank you very much for your help!
 A: $\newcommand{\+}{^{\dagger}}
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With $\ds{t \equiv \sin\pars{\theta}:}$
\begin{align}
\int_{0}^{\pi/2}{\dd\theta \over \root{\sin\pars{\theta}}}&
=\int_{0}^{1}t^{-1/2}\pars{1 - t^{2}}^{-1/2}\,\dd t
=\int_{0}^{1}t^{-1/4}\pars{1 - t}^{-1/2}\,\half\,t^{-1/2}\dd t
\\[3mm]&=\half\int_{0}^{1}t^{-3/4}\pars{1 - t}^{-1/2}\,\dd t
=\half\,{\rm B}\pars{{1 \over 4},\half}
\end{align}
where $\ds{{\rm B}\pars{x,y}}$ is the
Beta Function.

\begin{align}
\color{#00f}{\large\int_{0}^{\pi/2}{\dd\theta \over \root{\sin\pars{\theta}}}}&
=\half\,{\Gamma\pars{1/4}\Gamma\pars{1/2} \over \Gamma\pars{3/4}}
={\root{\pi} \over 2}\,{\Gamma\pars{1/4} \over \pi/\bracks{\Gamma\pars{1/4}\sin\pars{\pi/4}}}
\\[3mm]&={1 \over 2\root{\pi}}\,{\root{2} \over 2}\,\Gamma^{\,2}\pars{1 \over 4}
=\color{#00f}{\large{1 \over 4}\,\root{2 \over \pi}\Gamma^{\,2}\pars{1 \over 4}}
\approx 2.6221
\end{align}
  $\ds{\Gamma\pars{z}}$ is the Gamma Function and we used well known properties of it.

A: Put $x=\sin^2\theta$ then $\sqrt{\sin\theta}=x^{1/4}$ and $\theta=\arcsin\sqrt{x}$ hence
$$\eqalign{
\int_0^{\pi/2}\frac{d\theta}{\sqrt{\sin\theta}}&=\int_0^1\frac{1}{x^{1/4}}\frac{dx}{2\sqrt{x}\sqrt{1-x}}\cr
&=\frac{1}{2}\int_0^1x^{-3/4}(1-x)^{-1/2}dx\cr
&=\frac{1}{2}B\left(\frac{1}{4},\frac{1}{2}\right)=\frac{1}{2}\frac{\Gamma(1/4)\Gamma(1/2)}{\Gamma(3/4)}\cr
&=\frac{\sqrt{\pi}}{2}\frac{\Gamma^2(1/4)}{\Gamma(1/4)\Gamma(3/4)}\cr
&=\frac{\sqrt{\pi}}{2}\frac{\sin(\pi/4)}{\pi}\Gamma^2(1/4)\cr
&=\frac{1}{2\sqrt{2\pi}} \Gamma^2(1/4).
}
$$
wher $B$ and $\Gamma$ are the well-known Eulerian Functions.
A: In general we have 
$$2\int^{\pi/2}_0\sin^{2x-1}(\theta)\cos^{2y-1}(\theta)\,d\theta=B(x,y)$$
