Question

$\underset{0}{\overset{\mathrm{\pi }}{\int }}\log \left(1-\cos x\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let},I={\int }_0^{\mathrm{\pi }}\log \left(1-\cos x\right)\mathit{d}x \\ ={\int }_0^{\mathrm{\pi }}\log \left(2{\sin }^2\frac{x}{2}\right)dx \\ ={\int }_0^{\mathrm{\pi }}\log 2dx+2{\int }_0^{\mathrm{\pi }}\log \sin \frac{x}{2}dx \\ \mathrm{Let},t=\frac{x}{2}\mathrm{in}\mathrm{the}\mathrm{secong}\mathrm{integral}.\mathrm{then}dt=\frac{1}{2}dx \\ \mathrm{When}x\to 0;t\to 0\mathrm{and}x\to \mathrm{\pi };\mathrm{t}\to \frac{\mathrm{\pi }}{2} \\ I=\log 2{\left[x\right]}_0^{\mathrm{\pi }}+4{\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin tdt \\ =\mathrm{\pi log}2+4\times \left(-\frac{\mathrm{\pi }}{2}\log 2\right)\left[\mathrm{Where},{\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin \mathrm{t}\mathrm{dt}=-\frac{\mathrm{\pi }}{2}\log 2\right] \\ =-\mathrm{\pi }\log 2 \end{gathered}$$