$\int \sin 2 x \log _e \cos x d x$ is equal to

$\left(\frac{1}{2}-\log _e \cos x\right) \cos ^2 x+C$

Let

$I=\int \sin 2 x \log _e \cos x d x$

$\Rightarrow I=2 \int \cos x \log _e \cos x \sin x d x$

$\Rightarrow I=-2 \int t \log _e t d t$, where $t=\cos x$

$\Rightarrow I=-2\left\{\frac{t^2}{2} \log _e t-\frac{t^2}{4}\right\}+C=-t^2 \log _e t+\frac{t^2}{2}+C$

$\Rightarrow I=\left\{\frac{1}{2}-\log _e \cos x\right\} \cos ^2 x+C$