Prove without evaluating the integrals that:2∫π20xln(1−sinx)sinxdx=∫ππ2xln(1−sinx)sinxdx

Or equivalently:

∫π0xln(1−sinx)sinxdx=3∫π20xln(1−sinx)sinxdx

In contrast we have:

∫π0ln(1−sinx)sinxdx=2∫π20ln(1−sinx)sinxdx

This is of course easily provable by splitting the integral as ∫π20+∫ππ2 and letting x→π−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

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(1−sinx)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(π3−x)ln(1−sinx)sinxdx=0

I also noticed that we have:

J=∫ππ2xln(1−sinx)sinxdxx→π−x=∫π20(π−x)ln(1−sinx)sinxdx=πI(3π2)−I

⇒I+J=∫π0xln(1−sinx)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*