Question

The value of integral $\underset{0}{\overset{\pi /2}{\int }}(2\ln \sin x-\ln \sin 2x)dx$ is

• A. $\frac{\pi }{2}\left(\ln \frac{1}{2}\right)$
• B. $-\frac{\pi }{2}\left(\ln \frac{1}{2}\right)$
• C. $\frac{\pi }{4}\left(\ln \frac{1}{2}\right)$
• D. $-\frac{\pi }{4}\left(\ln \frac{1}{2}\right)$

Answer 1

The correct option is A $\frac{\pi }{2}\left(\ln \frac{1}{2}\right)$

Let $I=\underset{0}{\overset{\pi /2}{\int }}(2\ln \sin x-\ln \sin 2x)dx$

$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}(2\ln \sin x-\ln (2\sin x\cos x\left)\right\}dx$

$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}(2\ln \sin x-\ln \sin x-\ln \cos x-\ln 2)dx$

$\Rightarrow I=\underset{0}{\overset{\pi /2}{\int }}(\ln \sin x-\ln \cos x-\ln 2)dx\cdots \left(1\right)$

Using property, we get
$\left(\underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx\right)$

$\Rightarrow I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}(\ln \cos x-\ln \sin x-\ln 2)dx\cdots \left(2\right)$

Adding $\left(1\right)$ and $\left(2\right),$ we obtain

$2I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}(-\ln 2-\ln 2)dx$

$\Rightarrow 2I=-2\ln 2\underset{0}{\overset{\pi /2}{\int }}1\cdot dx$

$\Rightarrow I=-\ln 2\left[\frac{\pi }{2}\right]$

$\Rightarrow I=\frac{\pi }{2}(-\ln 2)$

$\Rightarrow I=\frac{\pi }{2}\left(\ln \frac{1}{2}\right)$