integration - How to evaluate $I=int_0^{pi/2}x^2ln(sin x)ln(cos x) mathrm dx$

Find the value of
$I=\displaystyle\int_0^{\pi/2}x^2\ln(\sin x)\ln(\cos x)\ \mathrm dx$

We have the information that
$J=\displaystyle\int_0^{\pi/2}x\ln(\sin x)\ln(\cos x)\ \mathrm dx=\dfrac{(\pi\ln{2})^2}{8}-\dfrac{\pi^4}{192}$

Answer

Tools Needed

$$\begin{align} \frac1{k(j-k)^2}&=\frac1{j^2k}-\frac1{j^2(k-j)}+\frac1{j(k-j)^2}\tag{1}\\ \frac1{k(j+k)^2}&=\frac1{j^2k}-\frac1{j^2(k+j)}-\frac1{j(k+j)^2}\tag{2}\\ \log(\sin(x))&=-\log(2)-\sum_{k=1}^\infty\frac{\cos(2kx)}{k}\tag{3}\\ \log(\cos(x))&=-\log(2)-\sum_{k=1}^\infty(-1)^k\frac{\cos(2kx)}{k}\tag{4}\\ \cos(2jx)\cos(2kx)&=\frac12\Big[\cos(2(j-k)x)+\cos(2(j+k)x)\Big]\tag{5}\\ \end{align}$$
$$\int_0^{\pi/2}x^2\cos(2kx)\,\mathrm{d}x=\left\{ \begin{array}{} (-1)^k\frac\pi{4k^2}&\text{if }k\ne0\\ \frac{\pi^3}{24}&\text{if }k=0 \end{array}\right.\tag{6}\\ $$

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Tool Use

$$\begin{align} &\int_0^{\pi/2}x^2\log(\sin(x))\log(\cos(x))\,\mathrm{d}x\\[12pt] &=\int_0^{\pi/2}x^2\left(\log(2)+\sum_{k=1}^\infty\frac{\cos(2kx)}{k}\right)\left(\log(2)+\sum_{k=1}^\infty(-1)^k\frac{\cos(2kx)}{k}\right)\,\mathrm{d}x\\[12pt] &=\log(2)^2\int_0^{\pi/2}x^2\,\mathrm{d}x +\log(2)\sum_{k=1}^\infty\frac1k\int_0^{\pi/2}x^2\cos(4kx)\,\mathrm{d}x\\ &+\sum_{j=1}^\infty\sum_{k=1}^\infty\frac{(-1)^k}{2jk}\int_0^{\pi/2}x^2\Big[\cos(2(j-k)x)+\cos(2(j+k)x)\Big]\,\mathrm{d}x\\[12pt] &=\frac{\pi^3}{24}\log(2)^2+\log(2)\frac\pi{16}\zeta(3)\\ &+\frac\pi8\sum_{j=1}^\infty\sum_{k=1}^\infty\frac{(-1)^j}{jk}\left[\mathrm{iif}\left(j=k,\frac{\pi^2}{6},\frac1{(j-k)^2}\right)+\frac1{(j+k)^2}\right]\\[12pt] &=\frac{\pi^3}{24}\log(2)^2+\log(2)\frac\pi{16}\zeta(3)\\ &+\frac\pi8\sum_{j=1}^\infty\frac{(-1)^j}{j}\sum_{k=1}^{j-1}\frac1{k(j-k)^2} +\frac\pi8\sum_{j=1}^\infty\frac{(-1)^j}{j^2}\frac{\pi^2}{6} +\frac\pi8\sum_{j=1}^\infty\frac{(-1)^j}{j}\sum_{k=j+1}^\infty\frac1{k(j-k)^2}\\ &+\frac\pi8\sum_{j=1}^\infty\frac{(-1)^j}{j}\sum_{k=1}^\infty\frac1{k(j+k)^2}\\[12pt] &=\frac{\pi^3}{24}\log(2)^2+\log(2)\frac\pi{16}\zeta(3)\\ &+\frac\pi8\sum_{j=1}^\infty\frac{(-1)^j}{j}\left(\frac2{j^2}H_{j-1}+\frac1jH_{j-1}^{(2)}\right) -\frac{\pi^5}{576} +\frac\pi8\sum_{j=1}^\infty\frac{(-1)^j}{j}\left(-\frac1{j^2}H_j+\frac1j\frac{\pi^2}{6}\right)\\ &+\frac\pi8\sum_{j=1}^\infty\frac{(-1)^j}{j}\left(\frac1{j^2}H_j-\frac1j\frac{\pi^2}{6}+\frac1jH_j^{(2)}\right)\\[12pt] &=\frac{\pi^3}{24}\log(2)^2+\log(2)\frac\pi{16}\zeta(3)\\ &+\frac\pi8\sum_{j=1}^\infty\frac{(-1)^j}{j}\left(\frac2{j^2}H_j+\frac2jH_j^{(2)}-\frac3{j^3}\right) -\frac{\pi^5}{576}\\[12pt] &=\frac{\pi^3}{24}\log(2)^2+\log(2)\frac\pi{16}\zeta(3)+\frac{11\pi^5}{5760} +\frac\pi4\sum(-1)^j\left(\frac1{j^3}H_j+\frac1{j^2}H_j^{(2)}\right)\\[12pt] &=\frac{\pi^3}{24}\log(2)^2+\log(2)\frac\pi{16}\zeta(3)-\frac{\pi^5}{960} -\frac\pi{16}\sum_{j=1}^\infty\frac1{j^3}H_{2j}\tag{7} \end{align}$$
Numerically, $(7)$ matches the integral. I'm working on the last harmonic sum. Both numerical integration and $(7)$ yield $0.0778219793722938643380944$.

Mathematica Help

Thanks to Artes' answer on Mathematica, I have verified that these agree to 100 places.