Question

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

Answer 1

$\begin{array}{c}I={\int }_0^{\frac{\pi }{2}}\frac{\sin 8x\log \left(co+x\right)}{\cos 2x}dx\to \left(i\right)\\ byapplyingpropertyofintegration\\ I={\int }_0^{a}faxdx={\int }_0^{a}f\left(a-x\right)dx\\ I={\int }_0^{\frac{\pi }{2}}\frac{\sin 8\left(\frac{\pi }{2}-x\right)\log \left[\cot \left(\frac{\pi }{2}-x\right)\right]}{\cos 2\left(\frac{\pi }{2}-x\right)}\\ I={\int }_0^{\frac{\pi }{2}}\frac{\sin \left(4x-8x\right)\log \left(\tan x\right)}{\cos \left(\pi -2x\right)}dx\\ I={\int }_0^{\frac{\pi }{2}}\frac{-\sin 8x\log \left(\frac{1}{\cot x}\right)}{-\cos 2x}dx\\ I={\int }_0^{\frac{\pi }{2}}\frac{-\sin 8x\log {\left(\cot x\right)}^{-1}}{-\cos 2x}\\ I={\int }_0^{\pi }-\frac{\sin 8x\log \left(\cot x\right)}{\cos 2x}\\ I={\int }_0^{\pi }\frac{\sin 8x\log \cot x}{\cos 2x}dx=-I\\ I=-I\\ 2I=0\\ I=0\end{array}$