∫ e cot x sin 2 x 2 log cosec x + sin 2 x d x =

$I=\int \frac{e^{\cot x}}{{\sin }^{2}x}\left(2\log cosecx+\sin 2x\right)dx$

$=\int e^{\cot x}{\mathrm{cosec}}^{2}x\left(2\log cosecx+\sin 2x\right)dx$

Let $\cot x=t; {\mathrm{cosec}}^{2}xdx=-dt$

$\Rightarrow I=-\int e^t\left(\log \left(1+t^2\right)+\frac{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$t$}\right.}}{1+{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$t^2$}\right.}}\right)dt$

$=-\int e^t\left(\log \left(1+t^2\right)+\frac{{\displaystyle 2t}}{1+{\displaystyle t^2}}\right)dt$

$=-e^t\log \left|1+t^2\right|+C$

$=-e^{{\cot }^{2}x}\log \left|1+{\cot }^{2}x\right|+C=-2e^{{\cot }^{2}x}\log \left|\mathrm{cosec}x\right|+C$