Question

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

Answer 1

${\int }_0^{\pi /2}\frac{\sin 8x\log \left(\cot x\right)}{\cos 2x}dx=I$ __(1)

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

${\int }_0^{\pi /2}\frac{\sin 8x\log \left(\tan x\right)dx}{\cos 2x}=I$ __(2)

adding 1 & 2

${\int }_0^{\pi /2}\frac{\sin 8x}{\cos 2x}\log \left(1\right)dx=2I$

$\therefore 2I=0$

$I=0$