# 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!

• It will require elliptic integrals. Are you familiar with those? – David H Apr 25 '14 at 17:43
• The title doesn't match the question. – Hans Lundmark Apr 25 '14 at 17:43
• @DavidH I know a little of elliptic integrals, not much – ziang chen Apr 25 '14 at 17:45
• @HansLundmark Thanks! – ziang chen Apr 25 '14 at 17:46
• – Lucian Apr 25 '14 at 18:30


\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.

• @DavidH That was unexpected. Thanks. – Felix Marin Apr 25 '14 at 17:56
• I guess I just liked how you managed it without elliptic integrals, and fairly quickly to boot. – David H Apr 25 '14 at 17:58

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.

In general we have

$$2\int^{\pi/2}_0\sin^{2x-1}(\theta)\cos^{2y-1}(\theta)\,d\theta=B(x,y)$$

• What is $B(x,y)$, the Beta function? – M Turgeon Apr 25 '14 at 23:16
• @MTurgeon, yes. – Zaid Alyafeai Apr 25 '14 at 23:19