Question

If ${\int }_0^{\pi /2}logsinxdx=k$, then ${\int }_0^{\pi }log(1+cosx)dx$ is given by.

Answer 1

We have

$\begin{array}{c}I={\int }_0^{\frac{\pi }{2}}\log \left(\sin 2x\right)dx=k\\ \\ I={\int }_0^{\pi }\log \left(1+\cos x\right)dx\\ I={\int }_0^{\pi }\log \left(1-\cos x\right)dx\\ 2I=2{\int }_0^{\frac{\pi }{2}}\log \left({\sin }^{2}x\right)dx\\ 2I=4{\int }_0^{\frac{\pi }{2}}\log \left(\sin x\right)dx\\ I=2{\int }_0^{\frac{\pi }{2}}\log \left(\sin x\right)dx\\ I=2k.\end{array}$