To solve the integral $\int_0^{\frac{\pi}{2}} e^{2x} \cos(3x) \, dx$, we will again use integration by parts.

Recall that integration by parts is given by: $\int u \, dv = uv - \int v \, du$

First, let's choose $u$ and $dv$: $u = e^{2x} \quad \text{and} \quad dv = \cos(3x) \, dx$

Now, compute $du$ and $v$: $du = 2e^{2x} \, dx \quad \text{and} \quad v = \frac{1}{3} \sin(3x)$

Applying integration by parts: $\int e^{2x} \cos(3x) \, dx = e^{2x} \cdot \frac{1}{3} \sin(3x) - \int \frac{1}{3} \sin(3x) \cdot 2e^{2x} \, dx$ $= \frac{1}{3} e^{2x} \sin(3x) - \frac{2}{3} \int e^{2x} \sin(3x) \, dx$

Now, we need to integrate $\int e^{2x} \sin(3x) \, dx$. We use integration by parts again for this integral. Let: $u = e^{2x} \quad \text{and} \quad dv = \sin(3x) \, dx$

Then: $du = 2e^{2x} \, dx \quad \text{and} \quad v = -\frac{1}{3} \cos(3x)$

Applying integration by parts again: $\int e^{2x} \sin(3x) \, dx = -\frac{1}{3} e^{2x} \cos(3x) - \int -\frac{1}{3} \cos(3x) \cdot 2e^{2x} \, dx$ $= -\frac{1}{3} e^{2x} \cos(3x) + \frac{2}{3} \int e^{2x} \cos(3x) \, dx$

Notice that we end up with the original integral $\int e^{2x} \cos(3x) \, dx$ on the right-hand side. Let's call this integral $I$: $I = \int e^{2x} \cos(3x) \, dx$ So we have: $I = \frac{1}{3} e^{2x} \sin(3x) - \frac{2}{3} \left( -\frac{1}{3} e^{2x} \cos(3x) + \frac{2}{3} \int e^{2x} \cos(3x) \, dx \right)$ $I = \frac{1}{3} e^{2x} \sin(3x) + \frac{2}{9} e^{2x} \cos(3x) - \frac{4}{9} I$

Now we solve for $I$: $I + \frac{4}{9} I = \frac{1}{3} e^{2x} \sin(3x) + \frac{2}{9} e^{2x} \cos(3x)$ $\frac{13}{9} I = \frac{1}{3} e^{2x} \sin(3x) + \frac{2}{9} e^{2x} \cos(3x)$ $I = \frac{9}{13} \left( \frac{1}{3} e^{2x} \sin(3x) + \frac{2}{9} e^{2x} \cos(3x) \right)$ $I = \frac{3}{13} e^{2x} \sin(3x) + \frac{2}{13} e^{2x} \cos(3x)$

Now we need to evaluate this expression from $0$ to $\frac{\pi}{2}$: $\left. \left( \frac{3}{13} e^{2x} \sin(3x) + \frac{2}{13} e^{2x} \cos(3x) \right) \right|_0^{\frac{\pi}{2}}$

Evaluating at the bounds: $= \left( \frac{3}{13} e^{2 \cdot \frac{\pi}{2}} \sin(3 \cdot \frac{\pi}{2}) + \frac{2}{13} e^{2 \cdot \frac{\pi}{2}} \cos(3 \cdot \frac{\pi}{2}) \right) - \left( \frac{3}{13} e^{0} \sin(0) + \frac{2}{13} e^{0} \cos(0) \right)$ $= \left( \frac{3}{13} e^{\pi} \sin\left(\frac{3\pi}{2}\right) + \frac{2}{13} e^{\pi} \cos\left(\frac{3\pi}{2}\right) \right) - \left( \frac{3}{13} \cdot 0 + \frac{2}{13} \cdot 1 \right)$ $= \left( \frac{3}{13} e^{\pi} \cdot (-1) + \frac{2}{13} e^{\pi} \cdot 0 \right) - \frac{2}{13}$ $= \left( -\frac{3}{13} e^{\pi} \right) - \frac{2}{13}$ $= -\frac{3}{13} e^{\pi} - \frac{2}{13}$ $= -\frac{1}{13} (3e^{\pi} + 2)$

So the value of the integral is: $\boxed{-\frac{1}{13} (3e^{\pi} + 2)}$

Would you like more details or have any questions?

Here are some related questions:

1. How do you integrate by parts?
2. What are some common mistakes when applying integration by parts?
3. How can we verify the result of an integral?
4. What are the properties of exponential functions?
5. How do trigonometric functions behave over different intervals?
6. What are some applications of definite integrals?
7. How can substitution simplify integrals?
8. What are some techniques for integrating products of functions?

Tip: When using integration by parts, choosing $u$ and $dv$ wisely can simplify the process significantly.