Question

Evaluate ${\int }_0^{\pi }\log (1-\cos x)dx$.

Answer 1

${\int }_0^{\pi }\log (1+\cos x)dx={\int }_0^{\pi /2}\log (1+\cos x)dx+{\int }_0^{\pi /2}\log (1+\cos (\pi -x\left)\right)dx$

$={\int }_0^{\pi /2}\log \left({\sin }^{2}x\right)dx=2{\int }_0^{\pi /2}\log \left(\sin x\right)dx$

$I={\int }_0^{\pi }\log \left(\sin x\right)dx=2{\int }_0^{\pi /2}\log \left(\sin 2t\right)dt=2{\int }_0^{\pi /2}\log \left(2\sin t\cos t\right)dt$

$=\pi \log 2+2{\int }_0^{\pi /2}\log \left(\sin t\right)dt+2{\int }_0^{\pi /2}\log \left(\cos t\right)dt=\pi \log 2+2I$

$\Rightarrow I=-\pi \log 2$

$\therefore {\int }_0^{\pi }log(1+cosx)dx=-\pi \log 2$