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.

7more comments