Prove, without evaluating the integrals that: $$\int_0^\pi\frac{x\ln(1-\sin x)}{\sin x}dx=3\int_0^\frac{\pi}{2}\frac{x\ln(1-\sin x)}{\sin x}dx$$

Originally I posted this here on MSE, however it's still unanswered.

As mentioned on the above link, both of the integrals are already evaluated independently (the integral from RHS can be found as seen here after substituting $\tan \frac{x}{2} \to x$), so I want to show this equality without calculating those integrals explicitly.

What seems quite interesting is that removing the $x$ from the numerator gives: $$\int_0^\pi\frac{\ln(1-\sin x)}{\sin x}dx=2\int_0^\frac{\pi}{2}\frac{\ln(1-\sin x)}{\sin x}dx$$ This can be shown quite easily without calculating any of the integrals, so that's what I seek to see by asking this question.

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    $\begingroup$ @StevenLandsburg I think what is meant is the analogous to the following: The last integral equation in the post can be seen by periodicity of $\dfrac{\ln(1-\sin(x))}{\sin(x)}$. Applying the periodicity of the function requires no actual evaluation of the integrals. Similarly, one could establish integral equalities just by a clever substitution, but not actually evaluating anything. I hope I captured the spirit here. $\endgroup$ Sep 9, 2021 at 14:53
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    $\begingroup$ @C.Caruvana Yes, that's what I'm looking for.// I'm not sure how to rewrite the first part in a better way. But mainly in the first link it's shown that $\int_0^\pi\frac{x\ln(1-\sin x)}{\sin x}dx=-\frac{3\pi^3}{8}$ and in the second linked post that $\int_0^\frac{\pi}{2} \frac{x\ln(1-\sin x)}{\sin x}dx=-\frac{\pi^3}{8}$ so the aim is to show this equality (but without using the results). $\endgroup$
    – Zacky
    Sep 9, 2021 at 15:04
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    $\begingroup$ @user64494 Pardon my density, but I'm not sure how to finish the argument based on this computation. It seems like what this establishes is that $\int_{0}^{\pi} \frac{x\ln(1-\sin(x))}{\sin(x)}\ dx = \frac{\pi}{2} \cdot \int_0^\pi \frac{\ln(1-\sin(x))}{\sin(x)}\ dx.$ How can I complete the desired equality from here? $\endgroup$ Sep 9, 2021 at 19:03
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    $\begingroup$ @StefanKohl - I don't understand your remark, I don't see votes to close - are there such? Despite some commenter remarks, this question seems nontrivial to me (certainly not a simple matter of symmetries). Possibly it won't have a good answer, but I don't think it's a trivial question. $\endgroup$ Sep 9, 2021 at 19:38
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    $\begingroup$ @MichaelEngelhardt The question was flagged for bounty abuse to prevent closure, by a user whom I believe can judge the question. Maybe he was mistaken anyway. While a question has a bounty, one cannot vote to close -- and the question lacked a top-level tag, which generally reduces visibility. I added such tag. If the question doesn't gather close votes now, the OP can simply start the bounty anew. $\endgroup$
    – Stefan Kohl
    Sep 9, 2021 at 19:46


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