Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$
$$\mathcal{J}=\int_{0}^{\pi/2}\frac{x^2}{\sin x}\,dx = \int_{0}^{1}\frac{\arcsin^2(x)}{x\sqrt{1-x^2}}\,dx=\sum_{n\geq 1}\frac{2^{2n-1}}{n^2\binom{2n}{n}}\int_{0}^{1}\frac{x^{2n-1}}{\sqrt{1-x^2}}\,dx \tag{1}$$
by the Maclaurin series of $\arcsin^2(x)$. Euler's Beta function then leads to
$$ \mathcal{J}=\sum_{n\geq 1}\frac{16^n}{4n^3 \binom{2n}{n}^2}=\phantom{}_4 F_3\left(1,1,1,1;\tfrac{3}{2},\tfrac{3}{2},2;1\right)\tag{2} $$
where the RHS is a manageable hypergeometric function (similar objects are evaluated both here and here) and as already shown by Claude Leibovici, $\mathcal{J}=4\int_{0}^{1}\frac{\arctan^2(u)}{u}\,du $ is simply given by a combination of a dilogarithm and a trilogarithm. Indeed
$$ \int_{0}^{\pi/2}\int_{0}^{\theta}\frac{u}{\sin u}\,du\,d\theta =-\pi G+\frac{7}{2}\zeta(3)\tag{3}$$
leading to $\mathcal{J}=2\pi G-\frac{7}{2}\zeta(3)$, has already been a key lemma in this historical thread.
An alternative way for proving this identity is just to write $\frac{x}{\sin x}$ and $|x|$ as Fourier cosine series.
The Shafer-Fink inequality leads to
$$ \int_{0}^{\pi/2}\frac{x^2}{\sin x}\,dx = 4 \int_{0}^{1}\frac{\arctan^2(u)}{u}\,du \approx \frac{6}{7}(3\sqrt{2}-5)+9\log\left(\frac{2\sqrt{2}+1}{3}\right)\approx 1.54.\tag{4}$$
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\begin{align}
\int_{0}^{\pi/2}{x^{2} \over \sin\pars{x}}\,\dd x & =
\left.\Re\int_{x\ =\ 0}^{x\ =\ \pi/2}{\bracks{-\ic\ln\pars{z}}^{2} \over
\pars{z - 1/z}/\pars{2\ic}}\,{\dd z \over \ic z}
\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\\[5mm] & =
\left.2\,\Re\int_{x\ =\ 0}^{x\ =\ \pi/2}{\ln^{2}\pars{z} \over 1 - z^{2}}\,\dd z
\,\right\vert_{\ z\ =\ \exp\pars{\ic x}}
\end{align}
$\ds{\ln}$ is the $\ds{\log}$-principal branch. Integration of $\ds{{\ln^{2}\pars{z} \over 1 - z^{2}}}$ along the path $\ds{C_{x}\cup C_{R}\cup C_{y}}$ vanishes out such that
$\ds{\int_{\large C_{R}}{\ln^{2}\pars{z} \over 1 - z^{2}}\,\dd z = -\int_{\large C_{y}}{\ln^{2}\pars{z} \over 1 - z^{2}}\,\dd z - \int_{\large C_{x}}{\ln^{2}\pars{z} \over 1 - z^{2}}\,\dd z}$
Then, \begin{align} \int_{0}^{\pi/2}{x^{2} \over \sin\pars{x}}\,\dd x &\ =\ \overbrace{-2\,\Re\int_{1}^{0}{\bracks{\ln\pars{y} + \pi\ic/2}^{\, 2} \over 1 + y^{2}}\,\ic\,\dd y}^{\ds{\mbox{along}\ C_{y}}}\ -\ \overbrace{2\,\Re\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x^{2}}\,\dd x} ^{\ds{\mbox{along}\ C_{x}}} \\[5mm] & = -2\pi\,\int_{0}^{1}{\ln\pars{y} \over 1 + y^{2}}\,\dd y - 2\int_{0}^{1}{\ln^{2}\pars{x} \over 1 - x^{2}}\,\dd x \end{align}
However, $\ds{\int_{0}^{1}{\ln\pars{y} \over 1 + y^{2}}\,\dd y = -G}$ where $\ds{G}$ is the Catalan Constant such that
\begin{align} \int_{0}^{\pi/2}{x^{2} \over \sin\pars{x}}\,\dd x & = 2\pi G - 2\sum_{n = 0}^{\infty}\ \overbrace{\int_{0}^{1}\ln^{2}\pars{x}x^{2n}\,\dd x} ^{\ds{2 \over \pars{2n + 1}^{3}}} \\[5mm] & = 2\pi G - 4\bracks{\sum_{n = 1}^{\infty}{1 \over n^{3}} - \sum_{n = 1}^{\infty}{1 \over \pars{2n}^{3}}} = 2\pi G - {7 \over 2}\sum_{n = 1}^{\infty}{1 \over n^{3}} \\[5mm] & = \bbx{2\pi G - {7 \over 2}\,\zeta\pars{3}} \approx 1.5480 \\ & \end{align}
We can adapt the formula derived in $(2)$ of this answer: $$ \log(2\cos(x/2))=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\cos(kx)\tag{1a} $$ Substituting $x\mapsto\pi-x$ in $\text{(1a)}$, we get $$ \log(2\sin(x/2))=\sum_{k=1}^\infty\frac{-1}k\cos(kx)\tag{1b} $$ Subtracting $\text{(1a)}$ from $\text{(1b)}$, the even terms cancel and we get $$ \bbox[5px,border:2px solid #C0A000]{\log(\tan(x/2))=\sum_{k=0}^\infty\frac{-2}{2k+1}\cos((2k+1)x)}\tag2 $$
Therefore,
$$
\begin{align}
\int_0^{\pi/2}\frac{x^2}{\sin(x)}\,\mathrm{d}x
&=\int_0^{\pi/2}x^2\,\mathrm{d}\log(\tan(x/2))\tag3\\
&=-2\int_0^{\pi/2}x\log(\tan(x/2))\,\mathrm{d}x\tag4\\
&=\sum_{k=0}^\infty\frac4{2k+1}\int_0^{\pi/2}x\cos((2k+1)x)\,\mathrm{d}x\tag5\\
&=\sum_{k=0}^\infty\frac4{(2k+1)^2}\int_0^{\pi/2}x\,\mathrm{d}\sin((2k+1)x)\tag6\\
&=\sum_{k=0}^\infty\frac4{(2k+1)^2}\left[x\sin((2k+1)x)+\frac{\cos((2k+1)x)}{2k+1}\right]_0^{\pi/2}\tag7\\
&=\sum_{k=0}^\infty\frac4{(2k+1)^2}\left[\frac\pi2(-1)^k-\frac1{2k+1}\right]\tag8\\
&=\bbox[5px,border:2px solid #C0A000]{2\pi\mathrm{G}-\frac72\zeta(3)}\tag9
\end{align}
$$
Explanation:
$(3)$: prepare to integrate by parts
$(4)$: integrate by parts
$(5)$: apply $(2)$
$(6)$: prepare to integrate by parts
$(7)$: integrate by parts
$(8)$: apply the limits of integration
$(9)$: evaluate, where $\mathrm{G}$ is Catalan's Constant