We need to find the indefinite integral of $e^{2x} \sin x$ with respect to $x$.

1. Problem Description

We need to find the indefinite integral of $e^{2x} \sin x$ with respect to $x$.

2. Solution Steps

We will use integration by parts twice. The formula for integration by parts is:

$\int u \, dv = uv - \int v \, du$

First, let $u = \sin x$ and $dv = e^{2x} dx$. Then $du = \cos x \, dx$ and $v = \int e^{2x} dx = \frac{1}{2}e^{2x}$.

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

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

Now, let's evaluate $\int e^{2x} \cos x \, dx$ using integration by parts again. Let $u = \cos x$ and $dv = e^{2x} dx$. Then $du = -\sin x \, dx$ and $v = \frac{1}{2}e^{2x}$.

$\int e^{2x} \cos x \, dx = \frac{1}{2} e^{2x} \cos x - \int \frac{1}{2} e^{2x} (-\sin x) \, dx$

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

Substitute this back into the first equation:

$\int e^{2x} \sin x \, dx = \frac{1}{2} e^{2x} \sin x - \frac{1}{2} \left( \frac{1}{2} e^{2x} \cos x + \frac{1}{2} \int e^{2x} \sin x \, dx \right)$

$\int e^{2x} \sin x \, dx = \frac{1}{2} e^{2x} \sin x - \frac{1}{4} e^{2x} \cos x - \frac{1}{4} \int e^{2x} \sin x \, dx$

Now, let $I = \int e^{2x} \sin x \, dx$.

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

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

$\frac{5}{4} I = \frac{1}{2} e^{2x} \sin x - \frac{1}{4} e^{2x} \cos x$

$I = \frac{4}{5} \left( \frac{1}{2} e^{2x} \sin x - \frac{1}{4} e^{2x} \cos x \right)$

$I = \frac{2}{5} e^{2x} \sin x - \frac{1}{5} e^{2x} \cos x$

$I = \frac{1}{5} e^{2x} (2 \sin x - \cos x) + C$

3. Final Answer

$\frac{1}{5} e^{2x} (2 \sin x - \cos x) + C$