Question

The value of ${\int }_0^{\pi /2}\frac{\sin 8x\log \left(\cot x\right)}{\cos 2x}dx$ is

• A. $0$
• B. $\pi$
• C. $\frac{5\pi }{2}$
• D. $\frac{3\pi }{2}$

Answer 1

The correct option is A $0$

Let , $I={\int }_0^{\pi /2}\frac{\sin 8x\log \cot x}{\cos 2x}dx$......(1)
By applying the property , ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
We get , $I={\int }_0^{\pi /2}\frac{\sin \left(8(\frac{\pi }{2}-x)\right)\log \cot (\frac{\pi }{2}-x)}{\cos 2(\frac{\pi }{2}-x)}dx$
$I={\int }_0^{\pi /2}\frac{\sin 8x\log \tan x}{\cos 2x}dx$.....(2)
$\left(1\right)+\left(2\right)\Rightarrow$ $2I={\int }_0^{\pi /2}\frac{\sin 8x\log \left(\tan x\cot x\right)}{\cos 2x}dx={\int }_0^{\pi /2}\frac{\sin 8x\log \left(1\right)}{\cos 2x}dx=0$
Therefore , $I=0$.....Ans