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

Recall that integration by parts is given by:
\[
\int u \, dv = uv - \int v \, du
\]
We need to choose \(u\) and \(dv\). Let's set:
\[
u = e^{2x} \quad \text{and} \quad dv = \cos(2x) \, dx
\]
Then we need to find \(du\) and \(v\):
\[
du = 2e^{2x} \, dx \quad \text{and} \quad v = \frac{1}{2} \sin(2x)
\]

Now we apply integration by parts:
\[
\int e^{2x} \cos(2x) \, dx = e^{2x} \cdot \frac{1}{2} \sin(2x) - \int \frac{1}{2} \sin(2x) \cdot 2e^{2x} \, dx
\]
\[
= \frac{1}{2} e^{2x} \sin(2x) - \int e^{2x} \sin(2x) \, dx
\]

Now we need to integrate \(\int e^{2x} \sin(2x) \, dx\). We use integration by parts again for this integral. Set:
\[
u = e^{2x} \quad \text{and} \quad dv = \sin(2x) \, dx
\]
Then:
\[
du = 2e^{2x} \, dx \quad \text{and} \quad v = -\frac{1}{2} \cos(2x)
\]

Now applying integration by parts again:
\[
\int e^{2x} \sin(2x) \, dx = -\frac{1}{2} e^{2x} \cos(2x) - \int -\frac{1}{2} \cos(2x) \cdot 2e^{2x} \, dx
\]
\[
= -\frac{1}{2} e^{2x} \cos(2x) + \int e^{2x} \cos(2x) \, dx
\]

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

Now we need to evaluate this expression from \(0\) to \(\frac{\pi}{2}\):
\[
\left. \frac{1}{4} e^{2x} (\sin(2x) + \cos(2x)) \right|_0^{\frac{\pi}{2}}
\]
Evaluating at the bounds:
\[
\left. \frac{1}{4} e^{2x} (\sin(2x) + \cos(2x)) \right|_0^{\frac{\pi}{2}} = \frac{1}{4} e^{2 \cdot \frac{\pi}{2}} (\sin(\pi) + \cos(\pi)) - \frac{1}{4} e^{2 \cdot 0} (\sin(0) + \cos(0))
\]
\[
= \frac{1}{4} e^{\pi} (0 + (-1)) - \frac{1}{4} (0 + 1)
\]
\[
= -\frac{1}{4} e^{\pi} - \frac{1}{4}
\]
\[
= -\frac{1}{4} (e^{\pi} + 1)
\]

So the value of the integral is:
\[
\boxed{-\frac{1}{4} (e^{\pi} + 1)}
\]

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.

Learn how to evaluate the integral of e^(2x) * cos(2x) from 0 to pi/2 using integration by parts. This problem involves exponential and trigonometric functions, requiring step-by-step application of integration techniques. Suitable for undergraduate students.

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