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

1. Problem Description

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

2. Solution Steps

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

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

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

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

Now we need to integrate $\int e^{2x} \cos{x} dx$. We will use integration by parts again.

Let $u = \cos{x}$ and $dv = e^{2x} dx$. Then $du = -\sin{x} dx$ and $v = \frac{1}{2} e^{2x}$.

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

Substituting this back into the first equation, we have:

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

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

Now, we solve for the integral:

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

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

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

$\int e^{2x} \sin{x} dx = \frac{2}{5} e^{2x} \sin{x} - \frac{1}{5} e^{2x} \cos{x} + C$

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

3. Final Answer

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