$\underset{0}{\overset{\pi /2}{\int }}\frac{\sin 8x.\log \left(\cot x\right)}{\cos 2x}dx=$

$\begin{array}{l}I=\underset{0}{\overset{\pi /2}{\int }}\frac{Sin8x.\log \left(\cot x\right)}{\cos 2x}dx\\ =\underset{0}{\overset{\pi /2}{\int }}\frac{\sin 8(\frac{\pi }{2}-x)\log \left(\cot (\frac{\pi }{2}-x)\right)}{Cos2(\frac{\pi }{2}-x)}dx\\ =\underset{0}{\overset{\pi /2}{\int }}\frac{(-\sin x)\left(\log \tan x\right)}{-\cos \left(2x\right)}dx\\ I=\underset{0}{\overset{\pi /2}{\int }}\frac{Sinx(-\log \cot x)}{\cos 2x}dx\\ =-I\\ \Rightarrow 2I=0\Rightarrow I=0\end{array}$