# 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

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ 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.

• @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