TITLE: Looking for another way to calculate this integral.
QUESTION [8 upvotes]: Calculate:

$$\mathcal{J}=\int_0^{\frac{\sqrt{3}}{2}}{\ln ^2x\frac{\arccos ^3\!\sqrt{1-x^2}}{\sqrt{1-x^2}}}\text{d}x$$


My attempt:
$$\int_0^{\frac{\sqrt{3}}{2}}{\ln ^2x\frac{\arccos ^3\!\sqrt{1-x^2}}{\sqrt{1-x^2}}}\text{d}x=\int_0^{\frac{\pi}{3}}{t^3\ln ^2\sin t\text{d}t=\frac{1}{16}}\int_0^{\frac{2\pi}{3}}{t^3\ln^2 \left( \sin \frac{t}{2} \right)}\text{d}t=\frac{1}{16}\mathcal{I}
\\
\begin{align*}
\mathcal{I}&=\int_0^{\frac{2\pi}{3}}{t^3\ln ^2\left( 2\sin \frac{t}{2} \right)}\text{d}t-2\ln 2\int_0^{\frac{2\pi}{3}}{t^3\ln \left( 2\sin \frac{t}{2} \right)}\text{d}t-\ln ^22\int_0^{\frac{2\pi}{3}}{t^3}\text{d}t
\\
&=-\text{Ls}_{6}^{\left( 3 \right)}\left( \frac{2\pi}{3} \right) +2\ln\text{2Ls}_{5}^{\left( 3 \right)}\left( \frac{2\pi}{3} \right) +\frac{4\pi ^4\ln ^22}{81}
\end{align*}
$$
Hence according to the formula
\begin{align*}
\zeta \left( n-k,\left\{ 1 \right\} \!^k \right) &-\sum_{j=0}^k{\frac{\left( -i\tau \right) \!^j}{j!}}\text{Li}_{2+k-j,\left\{ 1 \right\} ^{n-k-2}}\left( \text{e}^{i\tau} \right) \\
&=\frac{i^{k+1}\left( -1 \right) \!^{n-1}}{\left( n-1 \right) !}\sum_{r=0}^{n-k-1}{\sum_{m=0}^r{\left( \begin{array}{c}
 n-1\\
 k,m,r-m\\
\end{array} \right)}}\times \left( \frac{i}{2} \right) \!^r\left( -\pi \right) \!^{r-m}\text{Ls}_{n-\left( r-m \right)}^{\left( k+m \right)}\left( \tau \right)
\end{align*}
we have
$$\begin{align*}
\text{Ls}_{6}^{\left( 3 \right)}\left( \frac{2\pi}{3} \right) =&-\frac{946\pi ^6}{76545}-\frac{16\pi ^3}{27}\text{Gl}_{2,1}\left( \frac{2\pi}{3} \right) -\frac{8\pi ^2}{3}\text{Gl}_{3,1}\left( \frac{2\pi}{3} \right) +8\pi \text{Gl}_{4,1}\left( \frac{2\pi}{3} \right) \\
&+12\text{Gl}_{5,1}\left( \frac{2\pi}{3} \right) +6\zeta ^2\left( 3 \right)
\end{align*} 
\\
\text{Ls}_{5}^{\left( 3 \right)}\left( \frac{2\pi}{3} \right) =\frac{8\pi ^3}{27}\text{Cl}_2\left( \frac{2\pi}{3} \right) -4\pi \text{Cl}_4\left( \frac{2\pi}{3} \right) -\frac{16\pi ^2}{27}\zeta \left( 3 \right) +\frac{242}{27}\zeta \left( 5 \right) $$
and we get
\begin{align*}
\mathcal{I}=&\frac{946\pi ^6}{76545}+\frac{4\pi ^2\ln ^22}{81}+\frac{16\pi ^3}{27}\text{Gl}_{2,1}\left( \frac{2\pi}{3} \right) +\frac{8\pi ^2}{3}\text{Gl}_{3,1}\left( \frac{2\pi}{3} \right) -8\pi \text{Gl}_{4,1}\left( \frac{2\pi}{3} \right) \\
&-12\text{Gl}_{5,1}\left( \frac{2\pi}{3} \right) +\frac{16\ln 2\pi ^3}{27}\text{Cl}_2\left( \frac{2\pi}{3} \right) 
-8\ln 2\pi \text{Cl}_4\left( \frac{2\pi}{3} \right) -6\zeta ^2\left( 3 \right) \\
&-\frac{32\ln 2\pi ^2}{27}\zeta \left( 3 \right) +\frac{848\ln 2}{27}\zeta \left( 5 \right) 
\end{align*}
So
\begin{align*}
\mathcal{J}=\frac{1}{16}\mathcal{I}=&\frac{473\pi ^6}{612360}+\frac{\pi ^3}{27}\text{Gl}_{2,1}\left( \frac{2\pi}{3} \right) +\frac{\pi ^2}{6}\text{Gl}_{3,1}\left( \frac{2\pi}{3} \right) -\frac{\pi}{2}\text{Gl}_{4,1}\left( \frac{2\pi}{3} \right) 
\\
&-\frac{3}{4}\text{Gl}_{5,1}\left( \frac{2\pi}{3} \right) +\frac{\pi ^3\ln 2}{27}\text{Cl}_2\left( \frac{2\pi}{3} \right) -\frac{\pi \ln 2}{2}\text{Cl}_4\left( \frac{2\pi}{3} \right) +\frac{\pi ^4\ln ^22}{324}
\\
&-\frac{2\pi ^2\ln 2}{27}\zeta \left( 3 \right) -\frac{3}{8}\zeta ^2\left( 3 \right) +\frac{121\ln 2}{108}\zeta \left( 5 \right) 
\end{align*}
Notations:
\begin{align*}
&\text{Cl}_{a_1,...,a_k}\left( \theta \right) =\begin{cases}
 \Im \text{Li}_{a_1,...,a_k}\left( \text{e}^{i\theta} \right) \,\,\text{if}\,\,a_1+\cdots +a_k\,\,\text{even}\\
 \Re \text{Li}_{a_1,...,a_k}\left( \text{e}^{i\theta} \right) \,\,\text{if}\,\,a_1+\cdots +a_k\,\,\text{odd}\\
\end{cases}
\\
&\text{Gl}_{a_1,...,a_k}\left( \theta \right) =\begin{cases}
 \Re \text{Li}_{a_1,...,a_k}\left( \text{e}^{i\theta} \right) \,\,\text{if}\,\,a_1+\cdots +a_k\,\,\text{even}\\
 \Im \text{Li}_{a_1,...,a_k}\left( \text{e}^{i\theta} \right) \,\,\text{if}\,\,a_1+\cdots +a_k\,\,\text{odd}\\
\end{cases}
\\
&\text{Li}_{a_1,.\!\:.\!\:.\!\:,a_k}\left( z \right) =\sum_{n_1>\cdots >n_k>0}{\frac{z^{n_1}}{n_{1}^{a_1}\cdots n_{k}^{a_k}}}
\\
&\zeta \left( a_1,.\!\:.\!\:.\!\:a_k \right) =\text{Li}_{a_1,.\!\:.\!\:.\!\:,a_k}\left( 1 \right) 
\\
&\text{Ls}_{n}^{\left( k \right)}\left( \sigma \right) =-\int_0^{\sigma}{\theta ^k\ln ^{n-1-k}\left| 2\sin \frac{\theta}{2} \right|}\,\text{d}\theta
\end{align*}

So I wonder is there another way to calculate the integral,complex or real method.
Thanks!

REPLY [4 votes]: Integrating by parts
\begin{align}
 I&=\int_0^{\pi/3}t^3\ln^2\sin t\,dt\\
&=\frac{\pi^4}{324}\ln^2\frac{\sqrt{3}}{2}-\frac{1}{2}\int_0^{\pi/3}t^4\frac{\cos t}{\sin t}\ln \sin t\,dt\\
&=\frac{\pi^4}{324}\ln^2\frac{\sqrt{3}}{2}-\frac{1}{2}\int_0^{\sqrt{3}/2}\arcsin^4 u\ln u\frac{du}{u}
\end{align} 
In this paper series are given for the integer powers of arcsin function. In particular,
\begin{equation}
 \arcsin^4 u=\frac{3}{2}\sum_{k=2}^\infty\left\lbrace\sum_{m=1}^{k-1}\frac{1}{m^2} \right\rbrace\frac{\left( 2u \right)^{2k}}{\binom{2k}{k}k^2}
\end{equation} 
Then
\begin{align}
 I&=\frac{\pi^4}{324}\ln^2\frac{\sqrt{3}}{2}-\frac{3}{4}\sum_{k=2}^\infty\left\lbrace\sum_{m=1}^{k-1}\frac{1}{m^2} \right\rbrace\frac{2^{2k}}{\binom{2k}{k}k^2}
\int_0^{\sqrt{3}/2}u ^{2k-1}\ln u\,du\\
&=\frac{\pi^4}{324}\ln^2\frac{\sqrt{3}}{2}-\frac{3}{4}\sum_{k=2}^\infty\left\lbrace\sum_{m=1}^{k-1}\frac{1}{m^2} \right\rbrace\frac{2^{2k}}{\binom{2k}{k}k^2}\left[\frac{3^k}{2^{2k+1}k}\ln\frac{\sqrt{3}}{2}-\frac{3^k}{2^{2k+2}k^2}\right]\\
&=\frac{\pi^4}{324}\ln^2\frac{\sqrt{3}}{2}-\frac{3}{8}\sum_{k=2}^\infty\left\lbrace\sum_{m=1}^{k-1}\frac{1}{m^2} \right\rbrace\frac{3^k}{\binom{2k}{k}k^3}\left[\ln\frac{\sqrt{3}}{2}-\frac{1}{2k}\right]
\end{align}
We may derive an expression with a polygamma function as
\begin{equation}
 \sum_{m=1}^{k-1}\frac{1}{m^2}=\frac{\pi^2}{6}-\psi(1,k)
\end{equation} 
Other exponents for the arccos function can be treated in the same way, as general expressions for the integer powers of the arcsin function are given in the paper.