by HallsofIvy » Mon Jul 13, 2020 5:37 pm
That's weird. He starts by saying that he has already proved, I presume in a previous post, that [tex]\int_0^\pi ln(sin(x))dx= -2\pi[/tex]. Then he integrates [tex]\int_0^\pi ln(a sin(x))dx[/tex] by differentiating with respect to a to get a simpler integral where the constant can be related to [tex]\int_0^x ln(sin(x))dx[/tex].
I think it would be simpler to use the fact that [tex]ln(a sin(x))= ln(a)+ ln(sin(x))[/tex]. So the integral of ln(a sin(x)) is the integral of ln(a), a constant, which is [tex]ln(a)\pi[/tex] plus the integral of ln(sin(x) which was already known.