$\int \mathrm{e}^{2 x} \frac{(\sin 2 x \cos 2 x-1)}{\sin ^2 2 x} \mathrm{~d} x=$

To evaluate the integral $I = \int e^{2x} \frac{\sin 2x \cos 2x - 1}{\sin^2 2x} dx$, simplify the integrand first.

The numerator separates as $\frac{\sin 2x \cos 2x}{\sin^2 2x} - \frac{1}{\sin^2 2x} = \cot 2x - \csc^2 2x$, giving $I = \int e^{2x} (\cot 2x - \csc^2 2x) dx$.

Applying integration by parts to $\int e^{2x} \cot 2x dx$ with $u = \cot 2x$ and $dv = e^{2x} dx$ yields $\frac{1}{2} e^{2x} \cot 2x + \int e^{2x} \csc^2 2x dx$.

Substituting back, the $\int e^{2x} \csc^2 2x dx$ terms cancel, leaving $I = \frac{1}{2} e^{2x} \cot 2x + C$.

Alternatively, recognize that for $f(x) = \frac{1}{2} \cot 2x$, the integrand is $e^{2x}(2f(x) + f'(x))$, so by the standard formula, $I = e^{2x} f(x) + C = \frac{1}{2} e^{2x} \cot 2x + C$.

The result matches option D.