Definite Integral from 0 to Half Pi of Logarithm of Sine x

Theorem

 * $\displaystyle \int_0^{\pi/2} \map \ln {\sin x} \rd x = -\frac \pi 2 \ln 2$

Proof
By Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\sin x}$: Lemma, we have:


 * $\displaystyle \int_0^\pi \map \ln {\sin x} \rd x = 2 \int_0^{\pi/2} \map \ln {\sin x} \rd x$

We also have:

giving:

Therefore:


 * $\displaystyle \int_0^{\pi/2} \map \ln {\sin x} \rd x = -\frac \pi 2 \ln 2$