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

Question

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!

Answer 1

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)

Answer 2

Let $u=\sin x$, then $t=u^2$, and recognize the expression of the beta function in the new integral.

Answer 3

So, which answer is prefereable? $$ \sqrt{2}\;\mathrm{K}\left(\frac{1}{\sqrt{2}}\right) = \frac{1}{2}\;\mathrm{B}\left(\frac{1}{4},\frac{1}{2}\right) = \frac{\pi^{3/4}}{\sqrt{2}\,\Gamma(3/4)^2} $$

Answer 4

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{{\cal I} \equiv \int_{0}^{\pi/2}{\dd x \over \root{\sin\pars{x}}}:\ {\large ?}}$

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 $$ {\cal I}=\half\,{\rm B}\pars{{1 \over 4},\half}=\half\,{\Gamma\pars{1/4}\Gamma\pars{1/2} \over \Gamma\pars{3/4}} ={\root{\pi} \over 2}\,\Gamma^{2}\pars{1/4}\,{\sin\pars{\pi/4} \over \pi} $$ $$\color{#00f}{\large% \int_{0}^{\pi/2}{\dd x \over \root{\sin\pars{x}}} = {1 \over 4}\,\root{2 \over \pi}\,\Gamma^{\,2}\pars{1 \over 4}} \approx 2.62206 $$