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:2π20xln(1sinx)sinxdx=ππ2xln(1sinx)sinxdx

Or equivalently:
π0xln(1sinx)sinxdx=3π20xln(1sinx)sinxdx
In contrast we have:
π0ln(1sinx)sinxdx=2π20ln(1sinx)sinxdx
This is of course easily provable by splitting the integral as π20+ππ2 and letting xπx 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(1sinx)sinxdxtanx2x=210arctanxxln(1+x2(1x)2)dx=π38
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 (*):
I knew from here that:
I(3π2)=π20ln(1sinx)sinxdx=3π28
And since this result is very similar to the one from above, I tried to show that I=π3I(3π2), equivalent to:
π20(π3x)ln(1sinx)sinxdx=0
I also noticed that we have:
J=ππ2xln(1sinx)sinxdxxπx=π20(πx)ln(1sinx)sinxdx=πI(3π2)I
I+J=π0xln(1sinx)sinxdx=πI(3π2)=3π38
Of course now it’s trivial to deduce that 2I=J as we know the result for 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 (*) 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.

Answer

Attribution
Source : Link , Question Author : Zacky , Answer Author : Community

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