Integrate the function: $e^{2x} \sin x$

Let $I = \int e^{2x} \sin x \, dx$ ..........$(1)$
Integrating by parts,we use the formula $\int u \, dv = uv - \int v \, du$. Let $u = \sin x$ and $dv = e^{2x} \, dx$. Then $du = \cos x \, dx$ and $v = \frac{e^{2x}}{2}$.
$I = \sin x \cdot \frac{e^{2x}}{2} - \int \cos x \cdot \frac{e^{2x}}{2} \, dx$
$I = \frac{e^{2x} \sin x}{2} - \frac{1}{2} \int e^{2x} \cos x \, dx$
Again integrating by parts for $\int e^{2x} \cos x \, dx$,let $u = \cos x$ and $dv = e^{2x} \, dx$. Then $du = -\sin x \, dx$ and $v = \frac{e^{2x}}{2}$.
$I = \frac{e^{2x} \sin x}{2} - \frac{1}{2} \left[ \cos x \cdot \frac{e^{2x}}{2} - \int (-\sin x) \cdot \frac{e^{2x}}{2} \, dx \right]$
$I = \frac{e^{2x} \sin x}{2} - \frac{e^{2x} \cos x}{4} - \frac{1}{4} \int e^{2x} \sin x \, dx$
$I = \frac{e^{2x} \sin x}{2} - \frac{e^{2x} \cos x}{4} - \frac{1}{4} I$
$I + \frac{1}{4} I = \frac{e^{2x} \sin x}{2} - \frac{e^{2x} \cos x}{4}$
$\frac{5}{4} I = \frac{2e^{2x} \sin x - e^{2x} \cos x}{4}$
$I = \frac{e^{2x}}{5} (2 \sin x - \cos x) + C$,where $C$ is an arbitrary constant.