Question

Show that ${\int }_0^{\frac{\pi }{2}}[2\log \sin x-\log (\sin 2x\left)\right]dx=\frac{\pi }{2}{\log }_e\left(\frac{1}{2}\right)$.

Answer 1

Let $I_1={\int }_0^{\frac{\pi }{2}}(2\log \sin x-\log \sin 2x)dx$

$={\int }_0^{\frac{\pi }{2}}(2\log \sin x-\log 2\sin x\cos x)dx$

$={\int }_0^{\frac{\pi }{2}}(2\log \sin x-\log 2-\log \sin x-\log \cos x)dx$

$={\int }_0^{\frac{\pi }{2}}(\log \sin x-\log 2-\log \cos x)dx$

$={\int }_0^{\frac{\pi }{2}}\left(\log \sin x\right)dx-\log 2{\int }_0^{\frac{\pi }{2}}dx-{\int }_0^{\frac{\pi }{2}}\log \cos xdx$

Let $I_2={\int }_0^{\frac{\pi }{2}}\log \cos xdx$

$={\int }_0^{\frac{\pi }{2}}\log \cos (\frac{\pi }{2}-x)dx$

$={\int }_0^{\frac{\pi }{2}}\log \sin xdx$

Put the value of $I_2$ in $\left(1\right)$

$\therefore I_1={\int }_0^{\frac{\pi }{2}}\left(\log \sin x\right)dx-\log 2{\int }_0^{\frac{\pi }{2}}dx-{\int }_0^{\frac{\pi }{2}}\log \sin xdx$

$\Rightarrow I_1=-\log 2{\int }_0^{\frac{\pi }{2}}dx$

$=-\log 2{\left[x\right]}_0^{\frac{\pi }{2}}$

$=-\log 2[\frac{\pi }{2}-0]$

$=\frac{\pi }{2}\log \left(\frac{1}{2}\right)$