Prove without evaluating the integrals that:2∫π20xln(1−sinx)sinxdx=∫ππ2xln(1−sinx)sinxdx
In contrast we have:
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:
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:
And since this result is very similar to the one from above, I tried to show that I=π3I(3π2), equivalent to:
I also noticed that we have:
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.