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The integral provided is:

\[
\int e^{3\beta} \cos(4\beta) \, d\beta
\]

To solve this using integration by parts, let's use the method twice. Integration by parts is based on the formula:

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

### Step 1: First Application of Integration by Parts

Let:
\[
u = e^{3\beta}, \quad dv = \cos(4\beta) \, d\beta
\]

Then, we find:
\[
du = 3e^{3\beta} \, d\beta, \quad v = \frac{1}{4} \sin(4\beta)
\]

Now, apply the integration by parts formula:
\[
\int e^{3\beta} \cos(4\beta) \, d\beta = e^{3\beta} \cdot \frac{1}{4} \sin(4\beta) - \int \frac{1}{4} \sin(4\beta) \cdot 3e^{3\beta} \, d\beta
\]

Simplify the expression:
\[
= \frac{1}{4} e^{3\beta} \sin(4\beta) - \frac{3}{4} \int e^{3\beta} \sin(4\beta) \, d\beta
\]

### Step 2: Second Application of Integration by Parts

Now, we need to solve the integral \(\int e^{3\beta} \sin(4\beta) \, d\beta\).

Let:
\[
u = e^{3\beta}, \quad dv = \sin(4\beta) \, d\beta
\]

Then:
\[
du = 3e^{3\beta} \, d\beta, \quad v = -\frac{1}{4} \cos(4\beta)
\]

Applying the integration by parts formula again:
\[
\int e^{3\beta} \sin(4\beta) \, d\beta = -\frac{1}{4} e^{3\beta} \cos(4\beta) - \int -\frac{1}{4} \cos(4\beta) \cdot 3e^{3\beta} \, d\beta
\]

Simplify:
\[
= -\frac{1}{4} e^{3\beta} \cos(4\beta) + \frac{3}{4} \int e^{3\beta} \cos(4\beta) \, d\beta
\]

### Step 3: Solve for the Original Integral

Now, substitute this back into the equation from Step 1:

\[
\int e^{3\beta} \cos(4\beta) \, d\beta = \frac{1}{4} e^{3\beta} \sin(4\beta) - \frac{3}{4} \left[ -\frac{1}{4} e^{3\beta} \cos(4\beta) + \frac{3}{4} \int e^{3\beta} \cos(4\beta) \, d\beta \right]
\]

Simplify and solve for \(\int e^{3\beta} \cos(4\beta) \, d\beta\):

\[
\int e^{3\beta} \cos(4\beta) \, d\beta = \frac{1}{4} e^{3\beta} \sin(4\beta) + \frac{3}{16} e^{3\beta} \cos(4\beta) - \frac{9}{16} \int e^{3\beta} \cos(4\beta) \, d\beta
\]

Combine like terms:
\[
\left(1 + \frac{9}{16}\right) \int e^{3\beta} \cos(4\beta) \, d\beta = \frac{1}{4} e^{3\beta} \sin(4\beta) + \frac{3}{16} e^{3\beta} \cos(4\beta)
\]

This simplifies to:
\[
\frac{25}{16} \int e^{3\beta} \cos(4\beta) \, d\beta = \frac{1}{4} e^{3\beta} \sin(4\beta) + \frac{3}{16} e^{3\beta} \cos(4\beta)
\]

Finally, solve for the integral:
\[
\int e^{3\beta} \cos(4\beta) \, d\beta = \frac{1}{25} e^{3\beta} \left(4 \sin(4\beta) + 3 \cos(4\beta)\right) + C
\]

where \( C \) is the constant of integration.

Would you like more details on any part of this solution, or do you have any questions?

Here are some related questions:

1. How would you approach the integral \(\int e^{x} \cos(5x) \, dx\)?
2. What happens if you integrate \(\int e^{2\beta} \sin(3\beta) \, d\beta\) using the same method?
3. How can you check the result of an integration using differentiation?
4. What are the advantages of using integration by parts over other methods?
5. Can integration by parts be extended to multiple variables? If so, how?

**Tip:** When integrating products of exponential and trigonometric functions, integration by parts is usually the best approach to consider.

Learn how to integrate e^(3β) * cos(4β) using integration by parts with a detailed step-by-step solution. This problem covers integration techniques for exponential and trigonometric functions. Suitable for advanced college-level mathematics.

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