$$\int_{0}^{1}\ln({1-\sin x})\mathrm dx=\ln 2+\ln \pi$$
How may one show that?
I have try the following but it is not working.
using sub: $u=1-\sin x$
$$\int_{1}^{1-\sin 1}{\ln u\over u(2-u)}\mathrm du$$
$${2\over \ln 2}\int_{1}^{1-\sin 1}\left({1\over u}+{1\over 2-u}\right)\mathrm du$$
$${\ln 2\over 2}\ln\left({u\over 2-u}\right)$$
For $x∈(0,1]$, $1−\sin x<1$ and hence $\ln(1−\sin x)<0$. So, $\displaystyle \int_0^1\ln(1-\sin x)dx\le0$.
Since $\ln 2+\ln \pi>0$. The equality does not hold.