The integral $\int_0^{\frac{\pi}{2}}\left(\frac{x}{\sin x}\right)^2\text{d}x$ How to evaluate :
$$\int_0^{\frac{\pi}{2}}\left(\frac{x}{\sin x}\right)^2\text{d}x$$
I was wondering how would you use a series expansion? 
 A: Maple seems to get the answer using the Fundamental Theorem of Calculus, using the antiderivative
$$ {\frac {-2\,i{x}^{2}}{{{\rm e}^{2\,ix}}-1}}+2\,x\ln  \left( 1-{{\rm e}
^{ix}} \right) +2\,x\ln  \left( 1+{{\rm e}^{ix}} \right) -2\,i{x}^{2}-
2\,i\,{\rm polylog} \left( 2,{{\rm e}^{ix}} \right) -2\,i\,{\rm polylog}
 \left( 2,-{{\rm e}^{ix}} \right) 
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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\begin{align}
&\bbox[10px,#ffd]{\ds{\int_{0}^{\pi/2}\bracks{x \over \sin\pars{x}}^{2}\dd x}}  =
\left.\Re\int_{x\ =\ 0}^{x\ =\ \pi/2}{\bracks{-\ic\ln\pars{z}}^{\, 2} \over
\bracks{\pars{z - 1/z}/\pars{2\ic}}^{\, 2}}\,{\dd z \over \ic z}
\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] = &\
\left.4\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}{z\ln^{2}\pars{z} \over
\pars{1 - z^{2}}^{\, 2}}\,\dd z
\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}\label{1}\tag{1}
\end{align}

I'll evaluate the last expression by 'closing a contour' in the upper complex plane first quadrant. Namely, a quarter of an unit circle. \eqref{1} is the 'contribution' along $\ds{\expo{\ic\pars{0,\pi/2}}}$.

Then,
\begin{align}
&\bbox[10px,#ffd]{\ds{\int_{0}^{\pi/2}\bracks{x \over \sin\pars{x}}^{2}\dd x}}  =
\overbrace{-4\,\Im\int_{1}^{0}{\ic y\,\bracks{\ln\pars{y} + \pi\ic/2}^{\,2} \over \pars{1 + y^{2}}^{\, 2}}\,\ic\,\dd y}
^{\ds{\mbox{over the}\ y\mbox{-axis}: \pars{1,0}}}\ -\
\underbrace{4\,\ \overbrace{\Im\int_{0}^{1}{x\,\ln^{2}\pars{x} \over
\pars{1 - x^{2}}^{\, 2}}\,\dd x}^{\ds{=\ 0}}}
_{\ds{\mbox{over the}\ x\mbox{-axis}: \pars{0,1}}}
\\[5mm] & =
-4\pi\int_{0}^{1}{y\ln\pars{y} \over \pars{1 + y^{2}}^{\, 2}}\,\dd y
\,\,\,\stackrel{y^{2}\ \mapsto\ y}{=}\,\,\,
-\pi\int_{0}^{1}{\ln\pars{y} \over \pars{1 + y}^{\, 2}}\,\dd y
\\[5mm] & =
-\pi\sum_{n = 1}^{\infty}{-2 \choose n - 1}\
\underbrace{\int_{0}^{1}\ln\pars{y}\, y^{n - 1}\,\dd y}
_{\ds{-\,{1 \over n^{2}}}}\ =\
\pi\sum_{n = 1}^{\infty}{n \choose n - 1}\pars{-1}^{n - 1}\,{1 \over n^{2}} =
-\pi\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n}
\\[5mm] & = \bbx{\pi\ln\pars{2}} \approx 2.1776
\end{align}
A: Integrate $x^2 \csc^2(x)$ by parts and you get
$$2\int_0^{\pi \over 2} x\cot(x)$$
Do it again and you get
$$-2\int_0^{\pi \over 2} \ln(\sin x)$$
This is a famous integral (see the first answer in Computing the integral of $\log(\sin x)$) The result is
$$\pi \ln 2$$
A: Use Bernoulli's form of integration by parts formula 
$$\int udv = uv - u'v_1 + u''v_2 - ...$$
where $u',u''..$ are successive differentiation of the function $u(x)$ and $v_1, v_2 ..$ are successive integrals of the function $v(x)$.
We get 
$$\int_0^{\frac{\pi}{2}}\left(\frac{x}{\sin x}\right)^2\text{d}x = [-(x^2 \cot x)]_0^{\pi/2} + [2x \log(\sin x)]_0^{\pi/2} - 2 \int_0^{\frac{\pi}{2}} \log(\sin x) = \pi \log(2)$$
The first two integrals give 0 and the last integral is already computed in the post: Computing the integral of $\log(\sin x)$
A: $$\int_0^{\frac{\pi}{2}} \frac{x^2}{\sin^2 x}\, dx=\left[x^2\, (-\cot x)\right]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}} 2x\cot x\, dx = \Big[2x\log\sin x\Big]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} 2\log \sin x\, dx=\pi \log 2$$
