Csdrhrt

$$mathcal{J}=int_{0}^{pi/2}frac{x^2}{sin x},dx = int_{0}^{1}frac{arcsin^2(x)}{xsqrt{1-x^2}},dx=sum_{ngeq 1}frac{2^{2n-1}}{n^2binom{2n}{n}}int_{0}^{1}frac{x^{2n-1}}{sqrt{1-x^2}},dx tag{1}$$
by the Maclaurin series of $arcsin^2(x)$. Euler's Beta function then leads to
$$ mathcal{J}=sum_{ngeq 1}frac{16^n}{4n^3 binom{2n}{n}^2}=phantom{}_4 F_3left(1,1,1,1;tfrac{3}{2},tfrac{3}{2},2;1right)tag{2} $$
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}=4int_{0}^{1}frac{arctan^2(u)}{u},du $ is simply given by a combination of a dilogarithm and a trilogarithm. Indeed
$$ int_{0}^{pi/2}int_{0}^{theta}frac{u}{sin u},du,dtheta =-pi G+frac{7}{2}zeta(3)tag{3}$$
leading to $mathcal{J}=2pi G-frac{7}{2}zeta(3)$, 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
$$ int_{0}^{pi/2}frac{x^2}{sin x},dx = 4 int_{0}^{1}frac{arctan^2(u)}{u},du approx frac{6}{7}(3sqrt{2}-5)+9logleft(frac{2sqrt{2}+1}{3}right)approx 1.54.tag{4}$$

$begin{align}
J&=int_{0}^{pi/2}frac{x^2}{sin x},dx&
end{align}$

Perform the change of variable,

$displaystyle y=tanleft(frac{x}{2}right)$,

$begin{align}
J&=4int_0^1 frac{arctan^2 x}{x},dx\
&=4Big[ln xarctan ^2 xBig]_0^1-8int_0^1 frac{arctan xln x}{1+x^2},dx\
&=-8int_0^1 frac{arctan xln x}{1+x^2},dx\
end{align}$

For $xin [0;1]$, define $F$,

$begin{align} F(x)&=int_0^x frac{ln t}{1+t^2},dt\
&=int_0^1 frac{xln(xt)}{1+x^2t^2},dt
end{align}$

Observe that,

$displaystyle F(0)=0$ and, $displaystyle F(1)=-text{G}$.

$text{G}$ is the Catalan constant.

$begin{align}J&=-8Big[F(x)arctan xBig]_0^1+8int_0^1 int_0^1 frac{xln(tx)}{(1+t^2x^2)(1+x^2)},dt,dx\
&=2Gpi+8int_0^1 int_0^1 frac{xln x}{(1+t^2x^2)(1+x^2)},dt,dx+8int_0^1 int_0^1 frac{xln t}{(1+t^2x^2)(1+x^2)},dt,dx\
&=2Gpi+8int_0^1 Big[frac{arctan(tx)ln x}{1+x^2}Big]_{t=0}^{t=1},dx+4int_0^1 Big[frac{(ln(1+t^2x^2)-ln(1+x^2))ln t}{t^2-1}Big]_{x=0}^{x=1},dt\
&=2Gpi+8int_0^1 frac{arctan xln x}{1+x^2},dx+4int_0^1 frac{(ln(1+t^2)-ln 2)ln t }{t^2-1},dt\
&=2Gpi-J+4int_0^1 frac{(ln 2-ln(1+t^2))ln t }{1-t^2},dt\
end{align}$

Therefore,

$displaystyle J=text{G}pi+2int_0^1 frac{(ln 2-ln(1+x^2))ln x }{1-x^2},dx$

For $xin[0;1]$, define,

$begin{align}H(x)&=int_0^x frac{ln t}{1-t^2},dt\
&=int_0^1 frac{xln(tx)}{1-t^2x^2},dt\
end{align}$

Observe that,

$displaystyle H(0)=0$ and $displaystyle H(1)=-frac{pi^2}{8}$.

$begin{align}J&=text{G}pi+2Big[(ln 2-ln(1+x^2))H(x)Big]_0^1+4int_0^1int_0^1frac{x^2ln(tx)}{(1+x^2)(1-t^2x^2)},dt,dx\
&=text{G}pi+4int_0^1int_0^1frac{x^2ln(tx)}{(1+x^2)(1-t^2x^2)},dt,dx\
&=text{G}pi+4int_0^1int_0^1frac{x^2ln t}{(1+x^2)(1-t^2x^2)},dt,dx+4int_0^1int_0^1frac{x^2ln x}{(1+x^2)(1-t^2x^2)},dt,dx\
&=text{G}pi+4int_0^1Big[frac{ln t}{1+t^2}left(frac{ln(1+tx)}{2t}-frac{ln(1-tx)}{2t}-arctan xright)Big]_{x=0}^{x=1},dt+\
&2int_0^1 Big[frac{xln x}{1+x^2}lnleft(frac{1+tx}{1-tx}right)Big]_{t=0}^{t=1},dx\
&=text{G}pi+2int_0^1 frac{ln t}{t(1+t^2)}lnleft(frac{1+t}{1-t}right),dt-piint_0^1 frac{ln t}{1+t^2},dt+2int_0^1 frac{xln x}{1+x^2}lnleft(frac{1+x}{1-x}right),dx\
&=2text{G}pi+2int_0^1 frac{ln x}{x}lnleft(frac{1+x}{1-x}right),dx\
end{align}$

But, for $0leq x<1$,

$displaystyle frac{1}{x}lnleft(frac{1+x}{1-x}right)=2sum_{n=0}^{infty}frac{x^{2n}}{2n+1}$

Therefore,

$begin{align}int_0^1 frac{ln x}{x}lnleft(frac{1+x}{1-x}right),dx&=2int_0^1 left(sum_{n=0}^{infty}frac{x^{2n}}{2n+1}right)ln x,dx\
&=2 sum_{n=0}^{infty}int_0^1 frac{x^{2n}ln x}{2n+1},dx\
&=-2sum_{n=0}^{infty}frac{1}{(2n+1)^3}\
&=-2left(sum_{n=1}^{infty} frac{1}{n^3}-sum_{n=1}^{infty} frac{1}{(2n)^3}right)\
&=-2left(zeta(3)-frac{1}{8}zeta(3)right)\
&=-frac{7}{4}zeta(3)\
end{align}$

Therefore,

$ boxed{J=2text{G}pi-frac{7}{2}zeta(3)}$

Observe we have
begin{align}
I=int^{pi/2}_0 frac{x^2}{sin x} dx = int^{pi/2}_0 frac{x^2}{cosleft(frac{pi}{2}-x right)} dx = int^{pi/2}_0 frac{(frac{pi}{2}-u)^2}{cos u} du.
end{align}
Then using integration by parts, we see that
begin{align}
I&=left(frac{pi}{2}-uright)^2left{logleft|1 + sin uright|-log|cos u|right}bigg|^{pi/2}_0 + 2int^{pi/2}_0left(frac{pi}{2}-u right)log|sec u + tan u| du\
&= 2pi left(frac{1}{2}int_{0}^{pi/2}log|sec u+tan u| du right)-frac{7}{2}left(frac{4}{7}int^{pi/2}_0 u log|sec u+tan u| du right)\
&= 2pi G - frac{7}{2}zeta(3).
end{align}

Here, I have used the facts that
begin{align}
G= frac{1}{2}int_{0}^{pi/2}log|sec u+tan u| du
end{align}
and
begin{align}
zeta(3) = frac{4}{7}int^{pi/2}_0 u log|sec u+tan u| du.
end{align}
See here for reference.

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$

begin{align}
int_{0}^{pi/2}{x^{2} over sinpars{x}},dd x & =
left.Reint_{x = 0}^{x = pi/2}{bracks{-iclnpars{z}}^{2} over
pars{z - 1/z}/pars{2ic}},{dd z over ic z}
,rightvert_{ z = exppars{ic x}}
\[5mm] & =
left.2,Reint_{x = 0}^{x = pi/2}{ln^{2}pars{z} over 1 - z^{2}},dd z
,rightvert_{ z = exppars{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 sinpars{x}},dd x & =
overbrace{-2,Reint_{1}^{0}{bracks{lnpars{y} + piic/2}^{, 2} over
1 + y^{2}},ic,dd y}^{ds{mbox{along} C_{y}}} -
overbrace{2,Reint_{0}^{1}{ln^{2}pars{x} over 1 - x^{2}},dd x}
^{ds{mbox{along} C_{x}}}
\[5mm] & =
-2pi,int_{0}^{1}{lnpars{y} over 1 + y^{2}},dd y -
2int_{0}^{1}{ln^{2}pars{x} over 1 - x^{2}},dd x
end{align}

However, $ds{int_{0}^{1}{lnpars{y} over 1 + y^{2}},dd y = -G}$ where
$ds{G}$ is the Catalan Constant such that

begin{align}
int_{0}^{pi/2}{x^{2} over sinpars{x}},dd x & =
2pi G - 2sum_{n = 0}^{infty}
overbrace{int_{0}^{1}ln^{2}pars{x}x^{2n},dd x}
^{ds{2 over pars{2n + 1}^{3}}}
\[5mm] & =
2pi G - 4bracks{sum_{n = 1}^{infty}{1 over n^{3}} -
sum_{n = 1}^{infty}{1 over pars{2n}^{3}}} =
2pi G - {7 over 2}sum_{n = 1}^{infty}{1 over n^{3}}
\[5mm] & = bbx{2pi G - {7 over 2},zetapars{3}} approx 1.5480
end{align}

As pointed out within the other answers we want to prove that

$$mathfrak{I}=int_0^{pi/2}frac{x^2}{sin x}~dx=2pi G-frac72zeta(3)$$

As the OP showed $mathfrak{I}$ can be reduced to a linear combination of $x$ and the function $log(cot x)$

$$mathfrak{I}=int_0^{pi/2}frac{x^2}{sin x}~dx=8int_0^{pi/4}xlog(cot x)~dx$$

By applying the definition of the cotangent function followed up by the usage of the well-known Fourier series expansions of $log(cos x)$ and $log(sin x)$ this can be further simplified. Therefore we get

$$smallbegin{align}
mathfrak{I}=8int_0^{pi/4}xlog(cot x)~dx&=8left[int_0^{pi/4}xlog(cos x)~dx-int_0^{pi/4}xlog(sin x)~dxright]\
&=8left[int_0^{pi/4}xleft(-log(2)-sum_{n=1}^{infty}(-1)^nfrac{cos(2nx)}{n}right)~dx-int_0^{pi/4}xleft(-log(2)-sum_{n=1}^{infty}frac{cos(2nx)}{n}right)~dxright]\
&=8left[-sum_{n=1}^{infty}frac{(-1)^n}{n}underbrace{int_0^{pi/4}xcos(2nx)~dx}_I+sum_{n=1}^{infty}frac{1}{n}underbrace{int_0^{pi/4}xcos(2nx)~dx}_Iright]\
end{align}$$

The inner integral $I$ can be easily evaluated using IBP which leads to

$$I=int_0^{pi/4}xcos(2nx)~dx=frac{pi}{8n}sinleft(nfrac{pi}2right)-frac1{(2n)^2}$$

Plugging this into our original formula and followed by a little bit of algebraic manipulation we get

$$smallbegin{align}
mathfrak{I}&=8left[-sum_{n=1}^{infty}frac{(-1)^n}{n}left(frac{pi}{8n}sinleft(nfrac{pi}2right)-frac1{(2n)^2}right)+sum_{n=1}^{infty}frac{1}{n}left(frac{pi}{8n}sinleft(nfrac{pi}2right)-frac1{(2n)^2}right)right]\
&=2left[sum_{n=1}^{infty}frac{(-1)^n}{n^3}-sum_{n=1}^{infty}frac{1}{n^3}right]+pileft[sum_{n=1}^{infty}frac{1}{n^2}sinleft(nfrac{pi}2right)-sum_{n=1}^{infty}frac{(-1)^n}{n^2}sinleft(nfrac{pi}2right)right]
end{align}$$

The first terms can be evaluated in terms of the Riemann Zeta Function $zeta(s)$ and the Dirichlet Eta Function $eta(s)$ whereas for the second term we have to consider some basic properties of the Sine function. For $ninmathbb{N}>0$ the function $sinleft(nfrac{pi}2right)$ will be zero for all even $n$ and $-1$ and $1$ respectively for odd $n$ starting with $sinleft(frac{pi}2right)=1$ for $n=1$. Therefore all even terms vanish while the odd ones will remain with a oscillating negative sign. This leads to

$$smallbegin{align}
mathfrak{I}&=2left[sum_{n=1}^{infty}frac{(-1)^n}{n^3}-sum_{n=1}^{infty}frac{1}{n^3}right]+pileft[sum_{n=1}^{infty}frac{1}{n^2}sinleft(nfrac{pi}2right)-sum_{n=1}^{infty}frac{(-1)^n}{n^2}sinleft(nfrac{pi}2right)right]\
&=2[-eta(3)-zeta(3)]+pileft[sum_{n=0}^{infty}frac{1}{(2n+1)^2}(-1)^n-sum_{n=0}^{infty}frac{(-1)^{2n+1}}{(2n+1)^2}(-1)^nright]\
&=-2[(1-2^{-2})zeta(3)+zeta(3)]+2pisum_{n=0}^{infty}frac{(-1)^n}{(2n+1)^2}\
Leftrightarrowmathfrak{I}&=-frac72zeta(3)+2pi G
end{align}$$

where within the last step the functional relation between the Riemann Zeta Function and the Dirichlet Eta Function aswell as the series defintions of Catalan's Constant $G$ where used.

We can adapt the formula derived in $(2)$ of this answer:
$$
log(2cos(x/2))=sum_{k=1}^inftyfrac{(-1)^{k-1}}kcos(kx)tag{1a}
$$
Substituting $xmapstopi-x$ in $text{(1a)}$, we get
$$
log(2sin(x/2))=sum_{k=1}^inftyfrac{-1}kcos(kx)tag{1b}
$$
Subtracting $text{(1a)}$ from $text{(1b)}$, the even terms cancel and we get
$$
bbox[5px,border:2px solid #C0A000]{log(tan(x/2))=sum_{k=0}^inftyfrac{-2}{2k+1}cos((2k+1)x)}tag2
$$

Therefore,
$$
begin{align}
int_0^{pi/2}frac{x^2}{sin(x)},mathrm{d}x
&=int_0^{pi/2}x^2,mathrm{d}log(tan(x/2))tag3\
&=-2int_0^{pi/2}xlog(tan(x/2)),mathrm{d}xtag4\
&=sum_{k=0}^inftyfrac4{2k+1}int_0^{pi/2}xcos((2k+1)x),mathrm{d}xtag5\
&=sum_{k=0}^inftyfrac4{(2k+1)^2}int_0^{pi/2}x,mathrm{d}sin((2k+1)x)tag6\
&=sum_{k=0}^inftyfrac4{(2k+1)^2}left[xsin((2k+1)x)+frac{cos((2k+1)x)}{2k+1}right]_0^{pi/2}tag7\
&=sum_{k=0}^inftyfrac4{(2k+1)^2}left[fracpi2(-1)^k-frac1{2k+1}right]tag8\
&=bbox[5px,border:2px solid #C0A000]{2pimathrm{G}-frac72zeta(3)}tag9
end{align}
$$
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