Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\left(2\log \cos x-\log \sin 2x\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left(2\log \cos x-\log \sin 2x\right)dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[2\log \cos x-\log \left(2\sin x\cos x\right)\right]dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left(2\log \cos x-\log 2-\log \sin x-\log \cos x\right)dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\log \cos x-\log 2-\log \sin x\right)dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \cos xdx-{\int }_0^{\frac{\mathrm{\pi }}{2}}\log 2dx-{\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin xdx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \cos xdx-{\int }_0^{\frac{\mathrm{\pi }}{2}}\log 2dx-{\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin \left(\frac{\mathrm{\pi }}{2}-x\right)dx\left[\mathrm{Using}{\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right] \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \cos xdx-{\int }_0^{\frac{\mathrm{\pi }}{2}}\log 2dx-{\int }_0^{\frac{\mathrm{\pi }}{2}}\log \cos xdx \\ =-\log 2{\left[x\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ =-\frac{\mathrm{\pi }}{2}\log 2 \end{gathered}$$