[Mastermander]

\[
\int\limits_0^\pi {\ln \left( {1 + \cos x} \right)\,dx} = \int\limits_0^\pi {\ln \left( {1 - \cos x} \right)\,dx}
\]
\[
2\int\limits_0^\pi {\ln \left( {1 + \cos x} \right)\,dx} = \int\limits_0^\pi {\ln \left( {\sin ^2 x} \right)dx} = 2\int\limits_0^\pi {\ln \left( {\sin x} \right)dx}
\]
\[
\int\limits_0^\pi {\ln \left( {1 + \cos x} \right)\,dx} = \int\limits_0^\pi {\ln \left( {\sin x} \right)dx} \quad *
\]
\[
\int\limits_0^\pi {\ln \left( {\sin x} \right)dx} = \int\limits_0^{\pi /2} {\ln \left( {\sin x} \right)dx} + \int\limits_{\pi /2}^\pi {\ln \left( {\sin x} \right)dx}
\]
\[
f\left( x \right) \to f\left( {a - x} \right);\int\limits_{\pi /2}^\pi {\ln \left( {\sin x} \right)dx} = \int\limits_0^{\pi /2} {\ln \left( {\cos x} \right)dx} = \int\limits_0^{\pi /2} {\ln \left( {\sin x} \right)dx} = I
\]

\[
\int\limits_0^\pi {\ln \left( {\sin x} \right)dx} = 2\int\limits_0^{\pi /2} {\ln \left( {\sin x} \right)dx} = 2I\quad \quad **
\]
\[
I + I = \int\limits_0^{\pi /2} {\ln \left( {\sin x} \right)dx} + \int\limits_0^{\pi /2} {\ln \left( {\cos x} \right)dx} = \int\limits_0^{\pi /2} {\ln \left( {\sin x\cos x} \right)dx}
\]
\[
2I = \int\limits_0^{\pi /2} {\ln \left( {\frac{{\sin 2x}}{2}} \right)dx} = \int\limits_0^{\pi /2} {\ln \left( {\sin 2x} \right)dx} - \frac{\pi }{2}\ln 2
\]
\[
\int\limits_0^{\pi /2} {\ln \left( {\sin 2x} \right)dx} \quad ;\;2x = u \to dx = \frac{1}{2}du
\]
\[
\int\limits_0^{\pi /2} {\ln \left( {\sin 2x} \right)dx} = \frac{1}{2}\int\limits_0^\pi {\ln \left( {\sin u} \right)du} = \frac{1}{2}\left( {2I} \right) = I
\]
\[
2I = I - \frac{\pi }{2}\ln 2
\]
\[
I = - \frac{\pi }{2}\ln 2
\]
\[
\int\limits_0^\pi {\ln \left( {1-\cos x} \right)dx} = 2I = - \pi \ln 2\quad \quad \#
\]