The problem is to evaluate the indefinite integral $\int e^{2x} \sin x \, dx$.

1. Problem Description

The problem is to evaluate the indefinite integral $\int e^{2x} \sin x \, dx$.

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 $v = \int e^{2x} dx = \frac{1}{2}e^{2x}$.

Applying integration by parts, we get:

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

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

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

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

Applying integration by parts, we get:

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

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

Substitute this back into the first equation:

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

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

Now, we solve for $\int e^{2x} \sin x \, dx$.

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

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

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

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

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

3. Final Answer

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