Proving that ∫π0xln(1−sinx)sinxdx=3∫π20xln(1−sinx)sinxdx\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

Prove without evaluating the integrals that:$$2∫π20xln(1−sinx)sinxdx=∫ππ2xln(1−sinx)sinxdx2\int_0^\frac{\pi}{2}\frac{x\ln(1-\sin x)}{\sin x}dx=\int_\frac{\pi}{2}^\pi\frac{x\ln(1-\sin x)}{\sin x}dx\label{*}\tag{*}$$

Or equivalently:
$$∫π0xln(1−sinx)sinxdx=3∫π20xln(1−sinx)sinxdx\boxed{\int_0^\pi\frac{x\ln(1-\sin x)}{\sin x}dx=3\int_0^\frac{\pi}{2}\frac{x\ln(1-\sin x)}{\sin x}dx}$$
In contrast we have:
$$∫π0ln(1−sinx)sinxdx=2∫π20ln(1−sinx)sinxdx\boxed{\int_0^\pi\frac{\ln(1-\sin x)}{\sin x}dx=2\int_0^\frac{\pi}{2}\frac{\ln(1-\sin x)}{\sin x}dx}$$
This is of course easily provable by splitting the integral as $$∫π20+∫ππ2\int_0^\frac{\pi}{2}+\int_\frac{\pi}{2}^\pi$$ and letting $$x→π−xx\to \pi-x$$ in the second part, unfortunately this method doesn’t work for the other one.

I am already aware how to evaluate the integrals as we have:
$$I=∫π20xln(1−sinx)sinxdxtanx2→x=−2∫10arctanxxln(1+x2(1−x)2)dx=−π38\mathcal I= \int_0^\frac{\pi}{2}\frac{x\ln(1-\sin x)}{\sin x} dx\overset{\tan \frac{x}{2}\to x}=-2\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=-\frac{\pi^3}{8}$$
And the latter integral is evaluated in many ways here, so if you have other approaches please add them there.

Here’s how I came up with $$(*)\eqref{*}$$:
I knew from here that:
$$I(3π2)=∫π20ln(1−sinx)sinxdx=−3π28I\left(\frac{3\pi}{2}\right)=\int_0^\frac{\pi}{2}\frac{\ln(1-\sin x)}{\sin x}dx=-\frac{3\pi^2}{8}$$
And since this result is very similar to the one from above, I tried to show that $$I=π3I(3π2)\mathcal I=\frac{\pi}{3} I\left(\frac{3\pi}{2}\right)$$, equivalent to:
$$∫π20(π3−x)ln(1−sinx)sinxdx=0\boxed{\int_0^\frac{\pi}{2}\left(\frac{\pi}{3}-x\right)\frac{\ln(1-\sin x)}{\sin x}dx=0}$$
I also noticed that we have:
$$J=∫ππ2xln(1−sinx)sinxdxx→π−x=∫π20(π−x)ln(1−sinx)sinxdx=πI(3π2)−I\mathcal J=\int_\frac{\pi}{2}^\pi\frac{x\ln(1-\sin x)}{\sin x}dx\overset{x\to \pi-x}=\int_0^\frac{\pi}{2}\frac{(\pi-x)\ln(1-\sin x)}{\sin x}dx=\pi I\left(\frac{3\pi}{2}\right)-\mathcal I$$
$$⇒I+J=∫π0xln(1−sinx)sinxdx=πI(3π2)=−3π38\Rightarrow \mathcal I+\mathcal J=\int_0^\pi \frac{x\ln(1-\sin x)}{\sin x}dx=\pi I\left(\frac{3\pi}{2}\right)=-\frac{3\pi^3}{8}$$
Of course now it’s trivial to deduce that $$2I=J2\mathcal I=\mathcal J$$ as we know the result for $$I\mathcal I$$, but I’m interested to show that relationship without making use of the result or by calculating any of the integrals. If possible showing $$(*)\eqref{*}$$ using only integral manipulation (elementary tools such as substitution/integration by parts etc).
I hope there’s a nice slick way to do it as it will give an easy evaluation of the main integral.