Integration by using special functions

Question

$$\int ^{\pi }_{0}\dfrac {dt}{\sqrt {3-\cos t}}$$

How can you solve the following equation by using alpha/gamma functions and putting

$$\cos t=1-2\sqrt {u}$$

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} \color{#00f}{\large\int_{0}^{\pi}{\dd t \over \root{3 - \cos\pars{t}}}}&= \int_{0}^{\pi}{\dd t \over \root{3 - \bracks{1 - 2\sin^{2}\pars{t/2}}}} =\int_{0}^{\pi}{\dd t \over \root{2 + 2\sin^{2}\pars{t/2}}} \\[3mm]&=2\int_{0}^{\pi/2}{\dd t \over \root{2 + 2\sin^{2}\pars{t}}} =\root{2}\int_{0}^{\pi/2}{\dd t \over \root{1 + \sin^{2}\pars{t}}} \\[3mm]&=\color{#00f}{\large\root{2}\,{\rm K}\pars{-1}} \end{align} where $\ds{{\rm K}\pars{m} = \int_{0}^{\pi/2}{\dd t \over \root{1 - m\,\sin^{2}\pars{t}}}}$ is the Elliptic Integral of the First Kind.

Answer 2

Maple does this using EllipticK:

> int(1/sqrt(3-cos(t)),t=0..Pi); $$ \text{EllipticK}\left(\dfrac{1}{2}\sqrt{2}\right)$$

> select(has,[FunctionAdvisor(specialize,%)],GAMMA); $$ [[{\rm EllipticK} \left( \dfrac12\sqrt {2} \right) =\dfrac12{\frac {{\pi }^{ 3/2}}{ \Gamma \left( 3/4 \right) ^{2}}},\mbox { `with no restrictions`}]] $$