It is known that ∫0^π/2ln (sin x) dx=-(π)/(2)ln 2. Using this, ∫0^πln (1-cos x) dx is equal to

Use $1-\cos x=2\sin^2\dfrac x2$. Then $\ln(1-\cos x)=\ln2+2\ln\sin\dfrac x2$ and $\int_0^\pi\ln(1-\cos x)\,dx=\pi\ln2+2\int_0^\pi\ln\sin\dfrac x2\,dx$. Substitute $u=\dfrac x2$, so $dx=2\,du$ and the limits become $0$ to $\dfrac\pi2$: the last integral equals $2\int_0^{\pi/2}\ln\sin u\,du=2\left(-\dfrac\pi2\ln2\right)=-\pi\ln2$. Therefore the whole integral is $\pi\ln2+2(-\pi\ln2)=-\pi\ln2$. The integrand is singular at $x=0$ but integrable. Forgetting the factor $2$ from $dx=2\,du$ gives $-\dfrac{\pi}{2}\ln 2$ instead. Numerical integration of the original expression agrees with this closed form to many decimal places, which is a useful sanity check.