Question

If $I={\int }_0^{\pi /2}\frac{\sin 8x\log \left(\cot x\right)}{\cos 2x}dx$ then $I$ equals

• A. $-\pi /2$
• B. $-\pi /3$
• C. $-1/3$
• D. $0$

Answer 1

The correct option is D $0$

Using ${\int }_0^{\pi /2}f\left(x\right)dx={\int }_0^{\pi /2}f(a-x)dx,$
$I={\int }_0^{\pi /2}\frac{\sin (4\pi -8x)\log \cot (\pi /2-x)}{\cos (\pi -2x)}dx$
$I={\int }_0^{\pi /2}\frac{(\sin -8x)\left(\log \tan x\right)}{-\cos 2x}dx$
$=-I$ $[\because \log \tan x=-\log \cot x]$

$\Rightarrow 2I=0$ or $I=0$