Question

${\int }_0^{\pi /2}(2log\sin x-log\sin 2x)dx$ equals.

• A. $\pi log2$
• B. $-\pi log2$
• C. $\left(\pi /2\right)log2$
• D. $-\left(\pi /2\right)log2$

Answer 1

The correct option is D $-\left(\pi /2\right)log2$

We are given, $I={\int }_0^{\pi /2}(2\log \sin x-\log \sin 2x)dx$

$\to$ We know that

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$.

So, $I={\int }_0^{\pi /2}\left[2\log \right(\sin (\frac{\pi }{2}-x)\left)\right)-\log \left(\sin 2(\frac{\pi }{2}-2)\right)]dx$

$\therefore I={\int }_0^{\pi /2}(2\log \cos x-\log \sin 2x)dx........\left(2\right)$

We add Equation $\left(1\right)$ & $\left(2\right)$. We get

$\therefore 2I={\int }_0^{\pi /2}(2\log \sin x-\log \sin 2x+2\log \cos x-\log \sin x)dx$

$\therefore 2I={\int }_0^{\pi /2}[\log \sin x+\log \cos x-\log \sin 2x]dx$

So, $I={\int }_0^{\pi /2}[\log \sin x+\log \cos x-\log (2\sin x\cos x\left)\right]dx$

As we know, $\log (a.b)=\log a+\log b$

$I={\int }_0^{\pi /2}\log \sin x+\log \cos x-[\log 2+\log \sin x+\log \cos x]dx$

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

$I={\int }_0^{\pi /2}\log 2dx=-\log 2[x{]}_0^{\pi /2}$

$I=-\log 2[\frac{\pi }{2}-0]=-\left(\pi /2\right)\log 2$