Evaluating $\int_0^{\frac{\pi}{2}}x^2 \cot x\ln(1-\sin x)\mathrm{d}x$ I was able to find
$$\int_0^{\frac{\pi}{2}}x^2 \cot x\ln(1-\sin x)\mathrm{d}x=-\frac14\sum_{n=1}^\infty\frac{4^n}{{2n\choose n}}\frac{H_{2n}}{n^3}$$
$$=5\operatorname{Li}_4\left(\frac12\right)-\frac{65}{32}\zeta(4)-2\ln^2(2)\zeta(2)+\frac5{24}\ln^4(2)$$
by converting it to the sum above then evaluating this sum but many integrals and sums were involved in the calculations.
Do you have a different idea to find this integral or its sum?
 A: Here is one way to break up the integral
\begin{align}
I=\int_0^{\frac{\pi}{2}}x^2 \cot x\ln(1-\sin x)\ dx 
=I_1-2(I_2 -I_3)
\end{align}
where, with $\ln(1-\sin x)=\ln (\cos x) -2\tanh^{-1}(\tan\frac x2 )$ and $ \cot x = \csc x-\tan\frac x2$
\begin{align}
I_1=& \int_0^{\frac{\pi}{2}}x^2 \cot x \ln(\cos x)dx\\
=&\ \text{Li}_4(\frac12)-\frac{\pi^4}{720}-\frac{\pi^2}6\ln^22+\frac1{24}\ln^42 \\
\\
 I_2=&\int_0^{\frac{\pi}{2}}x^2 \csc x \tanh^{-1}(\tan\frac x2 ) \ dx \>\>\>\>\> t=\tan\frac x2\\
=& \ 4\int_0^1 \frac{(\tan^{-1}t)^2\tanh^{-1}t}{t}\  dt\\
=& \ 4\left(\pi \Im \text{Li}_3(\frac{1+i}2)+\frac{\pi}2G\ln2 -\frac{3\pi^4}{128} -\frac{\pi^2}{32}\ln^22\right)\\
\\
I_3=&\int_0^{\frac{\pi}{2}}x^2 \tan \frac x2 \ \tanh^{-1}(\tan\frac x2  )\ dx\\
=& \ 8 \int_0^1 \frac{t(\tan^{-1}t)^2\tanh^{-1}t}{1+t^2} dt\\
=& \ 8\bigg( \frac12\text{Li}_4(\frac12)+ \pi \Im \text{Li}_3(\frac{1+i}2)+\frac\pi2\ln2 G \\
&\hspace{20mm}-\frac{601\pi^4}{23040}-\frac{5\pi^2}{96}\ln^22+\frac1{48}\ln^42 \bigg)\\
\end{align}
The evaluation of the three integrals above are still involved, though familiar. Yet, as a by-product
$$\int_0^{\frac{\pi}{2}}x^2 \cot x\ln(1+\sin x)\ dx
=-3 \text{Li}_4(\frac12)+\frac{19\pi^4}{960}-\frac1{8}\ln^42 
$$
