Evaluate the following integrals:

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

Integrating by parts,

$I=\sin x \int e^{2 x} d x-\int \frac{d}{d x} \sin x \int e^{2 x} d x$

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

$I=\sin x \frac{e^{2 x}}{2}-\frac{1}{2} \int e^{2 x} \cos x d x$

Again integrating by parts,

$I=\sin x \frac{e^{2 x}}{2}-\frac{1}{2}\left\{\cos x \int e^{2 x} d x-\int \frac{d}{d x} \cos x \int e^{2 x} d x\right\}$

$I=\sin x \frac{e^{2 x}}{2}-\frac{1}{2}\left[\cos x \frac{e^{2 x}}{2}-\int(-\sin x) \frac{e^{2 x}}{2} d x\right]$

$I=\sin x \frac{e^{2 x}}{2}-\frac{1}{2}\left[\cos x \frac{e^{2 x}}{2}+\frac{1}{2} \int \sin x e^{2 x} d x\right]$

$I=\sin x \frac{e^{2 x}}{2}-\frac{1}{2} \cos x \frac{e^{2 x}}{2}-\frac{1}{4} I$

$I+\frac{I}{4}=\sin x \frac{e^{2 x}}{2}-\frac{1}{2} \cos x \frac{e^{2 x}}{2}$

$\frac{5}{4} I=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}$

$I=\frac{4}{5}\left[\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}\right]+c$

$I=\frac{e^{2 x}}{5}[2 \sin x-\cos x]+c$