Integration -- can't figure out the substitution I need to evaluate the integral
$$ \int_0^1 \frac{6\pi}{\lvert{2 - 3e^{2\pi i t}}\rvert^2} \,\mathrm{d}t.$$
The problem I am having is that I can't find a nice way to re-write the denominator to invoke a substitution. Perhaps I can view this as a contour integral in the complex plane? Any ideas appreciated.
 A: An idea:
$$2-3e^{2\pi it}=2-3\cos2\pi t-3i\sin2\pi t\implies |2-3e^{2\pi it}|^2=13-12\cos2\pi t\implies$$$${}$$
$$\int_0^1\frac{6\pi}{|2-3e^{2\pi it}|^2}dt=\int_0^1\frac{6\pi}{13-12\cos2\pi t}dt$$
You can now try some trigonometric substitution.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{1}{6\pi \over \verts{2 - 3\expo{2\pi\ic t}}^{2}}\,\dd t & =
{2\pi \over 3}\int_{0}^{1}
{\dd t\over \pars{1- 2\expo{-2\pi\ic t}/3}\pars{1- 2\expo{2\pi\ic t}/3}}
\\[5mm] & =
{2\pi \over 3}\sum_{k = 0}^{\infty}\sum_{j = 0}^{\infty}
\pars{2 \over 3}^{k + j}\
\overbrace{\int_{0}^{1}\expo{-2\pars{k - j}\pi\ic t}\,\dd t}^{\ds{\delta_{kj}}} =
{2\pi \over 3}\sum_{k = 0}^{\infty}\pars{4 \over 9}^{k}
\\[5mm] & =
{2\pi \over 3}\,{1 \over 1 - 4/9} = \bbx{\ds{6\pi \over 5}}
\end{align}
