We are asked to integrate \( e^{-3x} \cos(2x) \, dx \).

To solve this, we use the **integration by parts** technique twice, or we can recognize it as a standard case of solving by the **reduction formula** for integrals of the form \( e^{ax} \cos(bx) \, dx \).

Let's go through the process step by step using **integration by parts**:

### Step 1: Set up the integration by parts formula
We know the formula for integration by parts is:

\[
\int u \, dv = u v - \int v \, du
\]

We will use this twice. 

### Step 2: First integration by parts
Let:
- \( u = \cos(2x) \), so \( du = -2 \sin(2x) \, dx \)
- \( dv = e^{-3x} \, dx \), so \( v = \frac{e^{-3x}}{-3} \)

Now apply the integration by parts formula:

\[
\int e^{-3x} \cos(2x) \, dx = \frac{e^{-3x} \cos(2x)}{-3} - \int \left( \frac{e^{-3x} (-2) \sin(2x)}{-3} \right) \, dx
\]

Simplifying this:

\[
\int e^{-3x} \cos(2x) \, dx = \frac{e^{-3x} \cos(2x)}{-3} + \frac{2}{3} \int e^{-3x} \sin(2x) \, dx
\]

### Step 3: Second integration by parts
Now we integrate \( \int e^{-3x} \sin(2x) \, dx \) by parts.

Let:
- \( u = \sin(2x) \), so \( du = 2 \cos(2x) \, dx \)
- \( dv = e^{-3x} \, dx \), so \( v = \frac{e^{-3x}}{-3} \)

Now applying integration by parts again:

\[
\int e^{-3x} \sin(2x) \, dx = \frac{e^{-3x} \sin(2x)}{-3} - \int \left( \frac{e^{-3x} (2) \cos(2x)}{-3} \right) \, dx
\]

Simplifying this:

\[
\int e^{-3x} \sin(2x) \, dx = \frac{e^{-3x} \sin(2x)}{-3} - \frac{2}{3} \int e^{-3x} \cos(2x) \, dx
\]

### Step 4: Solving the system of equations
Now we substitute this result back into the original equation:

\[
\int e^{-3x} \cos(2x) \, dx = \frac{e^{-3x} \cos(2x)}{-3} + \frac{2}{3} \left( \frac{e^{-3x} \sin(2x)}{-3} - \frac{2}{3} \int e^{-3x} \cos(2x) \, dx \right)
\]

Simplifying:

\[
\int e^{-3x} \cos(2x) \, dx = \frac{e^{-3x} \cos(2x)}{-3} + \frac{2}{3} \cdot \frac{e^{-3x} \sin(2x)}{-3} - \frac{4}{9} \int e^{-3x} \cos(2x) \, dx
\]

### Step 5: Solve for the integral
Now we gather all terms involving \( \int e^{-3x} \cos(2x) \, dx \):

\[
\int e^{-3x} \cos(2x) \, dx + \frac{4}{9} \int e^{-3x} \cos(2x) \, dx = \frac{e^{-3x} \cos(2x)}{-3} + \frac{2}{9} e^{-3x} \sin(2x)
\]

\[
\left( 1 + \frac{4}{9} \right) \int e^{-3x} \cos(2x) \, dx = \frac{e^{-3x} \cos(2x)}{-3} + \frac{2}{9} e^{-3x} \sin(2x)
\]

\[
\frac{13}{9} \int e^{-3x} \cos(2x) \, dx = \frac{e^{-3x} \cos(2x)}{-3} + \frac{2}{9} e^{-3x} \sin(2x)
\]

Finally, solving for the integral:

\[
\int e^{-3x} \cos(2x) \, dx = \frac{9}{13} \left( \frac{e^{-3x} \cos(2x)}{-3} + \frac{2}{9} e^{-3x} \sin(2x) \right)
\]

\[
\int e^{-3x} \cos(2x) \, dx = \frac{e^{-3x}}{13} \left( -3 \cos(2x) - 2 \sin(2x) \right) + C
\]

Thus, the solution is:

\[
\boxed{\int e^{-3x} \cos(2x) \, dx = \frac{e^{-3x}}{13} (-3 \cos(2x) - 2 \sin(2x)) + C}
\]

Would you like more details or have any questions?

Here are 5 related questions:
1. How does the reduction formula work for integrals like this?
2. What are the key conditions for applying integration by parts?
3. How would you approach the integral \( e^{ax} \sin(bx) \, dx \)?
4. Why do we use integration by parts twice in this problem?
5. Can you explain the process for solving systems of integrals?

**Tip:** When solving complex integrals, organizing terms carefully helps avoid mistakes when performing multiple integration steps.

Learn how to solve the integral ∫ e^-3x cos(2x) dx using the integration by parts technique. This step-by-step guide covers key calculus concepts including exponential and trigonometric functions, and the reduction formula. Suitable for college-level calculus students.

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