Looking for another way to calculate this integral.

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.