I know the integral is possible with a simple fourier series expansion of $-\log(\sin(x))$
But I am interested in complex analysis, so I want to try this.
$$I = \int_{0}^{\pi} \log(\sin(x)) dx$$
The substitution $x = \arcsin(t)$ first comes to mind.
But that substitution isnt valid as,
The upper and lower bounds would both be $0$ because $\sin(\pi) = \sin(0) = 0$
What is a workaround using inverse trig or some other way?
Inverse sine is good, because that gives us a denominator from which we will be able to find poles.
$$x = arctan(t)$$
Is also good, but it would hard to do.
Idea?