Question

Prove that:
${\int }_0^{\frac{\pi }{2}}(2\log \sin x-\log \sin 2x)dx$

Answer 1

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

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

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

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

$={\int }_0^{\frac{\pi }{2}}\log \sin xdx-{\int }_0^{\frac{\pi }{2}}\log 2dx-{\int }_0^{\frac{\pi }{2}}\log \cos xdx$ ........$\left(1\right)$

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

using the property, ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$

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

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

$\therefore I_1={\int }_0^{\frac{\pi }{2}}\log \sin xdx-{\int }_0^{\frac{\pi }{2}}\log 2dx-{\int }_0^{\frac{\pi }{2}}\log \sin xdx$

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

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

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

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