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We are asked to use integration by parts to solve the integral

\[
I = \int e^x \cos(2x) \, dx
\]

### Step 1: Set up integration by parts
Recall the formula for integration by parts:

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

We need to select parts for \(u\) and \(dv\). Let:
- \( u = e^x \), so \( du = e^x \, dx \)
- \( dv = \cos(2x) \, dx \), so \( v = \frac{1}{2} \sin(2x) \) (since the integral of \(\cos(2x)\) is \(\frac{1}{2} \sin(2x)\))

### Step 2: Apply integration by parts
Using the formula for integration by parts:

\[
I = e^x \cdot \frac{1}{2} \sin(2x) - \int \frac{1}{2} \sin(2x) \cdot e^x \, dx
\]

This simplifies to:

\[
I = \frac{e^x}{2} \sin(2x) - \frac{1}{2} \int e^x \sin(2x) \, dx
\]

### Step 3: Solve the new integral
We now need to compute \( \int e^x \sin(2x) \, dx \). Let this integral be \( J \), so:

\[
J = \int e^x \sin(2x) \, dx
\]

We apply integration by parts again. Let:
- \( u = e^x \), so \( du = e^x \, dx \)
- \( dv = \sin(2x) \, dx \), so \( v = -\frac{1}{2} \cos(2x) \)

Thus:

\[
J = e^x \cdot \left( -\frac{1}{2} \cos(2x) \right) - \int -\frac{1}{2} \cos(2x) \cdot e^x \, dx
\]

Simplifying:

\[
J = -\frac{e^x}{2} \cos(2x) + \frac{1}{2} \int e^x \cos(2x) \, dx
\]

### Step 4: Solve for \( I \)
Notice that the integral on the right side of the equation for \( J \) is the original integral \( I \). Thus:

\[
J = -\frac{e^x}{2} \cos(2x) + \frac{1}{2} I
\]

Substitute this into the equation for \( I \):

\[
I = \frac{e^x}{2} \sin(2x) - \frac{1}{2} \left( -\frac{e^x}{2} \cos(2x) + \frac{1}{2} I \right)
\]

Simplify:

\[
I = \frac{e^x}{2} \sin(2x) + \frac{e^x}{4} \cos(2x) - \frac{1}{4} I
\]

### Step 5: Solve for \( I \)
Now, collect terms involving \( I \):

\[
I + \frac{1}{4} I = \frac{e^x}{2} \sin(2x) + \frac{e^x}{4} \cos(2x)
\]

This simplifies to:

\[
\frac{5}{4} I = \frac{e^x}{2} \sin(2x) + \frac{e^x}{4} \cos(2x)
\]

Multiply both sides by \( \frac{4}{5} \) to solve for \( I \):

\[
I = \frac{2e^x}{5} \sin(2x) + \frac{e^x}{5} \cos(2x)
\]

Thus, we have shown that:

\[
\int e^x \cos(2x) \, dx = \frac{2e^x}{5} \sin(2x) + \frac{e^x}{5} \cos(2x) + C
\]

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

---

Would you like further details on any specific step, or do you have any questions?

### Related Questions:
1. How does integration by parts work for products of exponential and trigonometric functions in general?
2. What is the derivation of the integration formula for \( \cos(2x) \)?
3. Can we use substitution instead of integration by parts for similar integrals?
4. How do the signs in trigonometric integrals like \( \cos(2x) \) and \( \sin(2x) \) affect the integration process?
5. What are common mistakes made when solving integrals using integration by parts?

**Tip:** Always double-check which part of the product you are choosing as \( u \) and \( dv \) to simplify the resulting integral.

Use integration by parts to show that ∫ e^x cos(2x) dx = (2e^x / 5) sin(2x) + (e^x / 5) cos(2x) + c, c ∈ ℝ.

Step-by-step solution using integration by parts to evaluate the integral ∫ e^x cos(2x) dx. Learn how to solve exponential and trigonometric integrals with detailed breakdowns of key steps, including applications of integration by parts and solving for unknown integrals.

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