Question

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

• A. $-\frac{\pi }{2}log\frac{1}{2}$
• B. $\frac{\pi }{2}log\frac{1}{2}$
• C. $\frac{\pi }{2}log2$
• D. $-\pi log\frac{1}{2}$

Answer 1

The correct option is B $\frac{\pi }{2}log\frac{1}{2}$

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

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

$I=$${\int }_0^{\pi /2}\log \tan xdx$$+{\int }_0^{\pi /2}\log \frac{1}{2}dx$$={\int }_0^{\pi /2}\log \tan xdx$$+\frac{\pi }{2}\log \frac{1}{2}dx$........(1).

$I=$${\int }_0^{\pi /2}\log \tan (\frac{\pi }{2}-x)dx+\frac{\pi }{2}\log \frac{1}{2}$ [ Using properties of definite integral]

$I=$${\int }_0^{\pi /2}\log \cot xdx+\frac{\pi }{2}\log \frac{1}{2}$ .......(2).

Now adding (1) and (2) we get,

$2I=$${\int }_0^{\pi /2}\log 1dx+\pi \log \frac{1}{2}$

$2I=\pi \log \frac{1}{2}$

$I=\frac{\pi }{2}\log \frac{1}{2}$.