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.