Question

By using the properties of definite integrals, evaluate the integral ${\int }_0^{\frac{\pi }{2}}(2\log \sin x-\log \sin 2x)dx$

Answer 1

Let $I={\int }_0^{\frac{\pi }{2}}(2\log \sin x-\log \sin 2x)dx$
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\{2\log \sin x-\log (2\sin x\cos x\left)\right\}dx$
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\{2\log \sin x-\log \sin x-\log \cos x-\log 2\}dx$
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\{\log \sin x-\log \cos x-\log 2\}dx$ ............ (1)
It is known that, $\left({\int }_o^{a}f\right(x)dx={\int }_0^{a}f(a-x\left)dx\right)$
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\{\log \cos x-\log \sin x-\log 2\}dx$ ........... (2)
Adding (1) and (2), we obtain
$2I={\int }_0^{\frac{\pi }{2}}(-\log 2-\log 2)dx$
$\Rightarrow 2I=-2\log 2{\int }_0^{\frac{\pi }{2}}1\cdot dx$
$I=-\log 2\left[\frac{\pi }{2}\right]$
$\Rightarrow I=\frac{\pi }{2}(-\log 2)$
$\Rightarrow I=\frac{\pi }{2}\left[\log \frac{1}{2}\right]$
$\Rightarrow I=\frac{\pi }{2}\log \frac{1}{2}$