Evaluate the following integrals:

Let $I=\int e^{2 x} \sin x \cos x d x$

$=\frac{1}{2} \int \mathrm{e}^{2 \mathrm{x}} 2 \sin \mathrm{x} \cos \mathrm{x} \mathrm{dx}$

$=\frac{1}{2} \int \mathrm{e}^{2 \mathrm{x}} \sin 2 \mathrm{x} \mathrm{dx}$

We know that,

$\int e^{a x} \sin b x d x=\frac{e^{a x}}{a^{2}+b^{2}}\{a \sin b x-b \cos b x\}+c$

$=\frac{e^{2 x}}{8}\{2 \sin 2 x-2 \cos 2 x\}+c$

$I=\frac{1}{2} \frac{e^{2 x}}{8}\{2 \sin 2 x-2 \cos 2 x\}+c$

$I=\frac{e^{2 x}}{8}\{\sin 2 x-\cos 2 x\}+c$