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Let's solve each of the three integrals step by step.

### Problem 1: \( \int_0^\pi x \sin x \cos x \, dx \)
This can be simplified using the trigonometric identity:
\[
\sin x \cos x = \frac{1}{2} \sin(2x)
\]
Thus, the integral becomes:
\[
\int_0^\pi x \sin x \cos x \, dx = \frac{1}{2} \int_0^\pi x \sin(2x) \, dx
\]

We can use integration by parts, where:
- Let \( u = x \) and \( dv = \sin(2x) dx \),
- So, \( du = dx \) and \( v = -\frac{1}{2} \cos(2x) \).

Now applying the integration by parts formula:
\[
\int u \, dv = uv - \int v \, du
\]
\[
\frac{1}{2} \left[ -\frac{x}{2} \cos(2x) \right]_0^\pi + \frac{1}{2} \int_0^\pi \frac{1}{2} \cos(2x) \, dx
\]
Evaluating this integral gives:
\[
I_1 = \boxed{0}
\]

### Problem 2: \( \int_{-1}^{1} |x| e^x \, dx \)
The absolute value function splits the integral into two parts:
\[
\int_{-1}^{1} |x| e^x \, dx = \int_{-1}^{0} (-x) e^x \, dx + \int_0^1 x e^x \, dx
\]

Now, compute each part:
1. For \( \int_{-1}^{0} (-x) e^x \, dx \), we use integration by parts:
   - Let \( u = -x \) and \( dv = e^x dx \),
   - So, \( du = -dx \) and \( v = e^x \),
   \[
   \int_{-1}^{0} (-x) e^x \, dx = -[(-x) e^x]_ {-1}^{0} + \int_{-1}^{0} e^x \, dx
   \]
   Evaluating gives:
   \[
   \int_{-1}^{0} (-x) e^x \, dx = e^{-1}
   \]

2. For \( \int_0^1 x e^x \, dx \), we use the same process:
   \[
   \int_0^1 x e^x \, dx = [xe^x]_0^1 - \int_0^1 e^x \, dx = e - 1
   \]

Thus, the final result is:
\[
I_2 = e - \frac{1}{e} - 1 = \boxed{e - 1 - \frac{1}{e}}
\]

### Problem 3: \( \int_{\frac{1}{e}}^e |\ln x| \, dx \)
We need to split the integral based on the behavior of \( |\ln x| \):
- For \( x \in \left[\frac{1}{e}, 1\right] \), \( \ln x \leq 0 \), so \( |\ln x| = -\ln x \),
- For \( x \in [1, e] \), \( \ln x \geq 0 \), so \( |\ln x| = \ln x \).

Thus, the integral becomes:
\[
\int_{\frac{1}{e}}^e |\ln x| \, dx = \int_{\frac{1}{e}}^1 (-\ln x) \, dx + \int_1^e \ln x \, dx
\]

Now compute each part:
1. For \( \int_{\frac{1}{e}}^1 (-\ln x) \, dx \), we use the formula \( \int \ln x \, dx = x \ln x - x \):
   \[
   \int_{\frac{1}{e}}^1 (-\ln x) \, dx = \left[ -x \ln x + x \right]_{\frac{1}{e}}^1 = 1 - \frac{1}{e}
   \]

2. For \( \int_1^e \ln x \, dx \), the same formula gives:
   \[
   \int_1^e \ln x \, dx = \left[ x \ln x - x \right]_1^e = e - 1
   \]

Thus, the final result is:
\[
I_3 = 1 - \frac{1}{e} + e - 1 = \boxed{e - \frac{1}{e}}
\]

---

Would you like further details or clarifications? Here are some related questions to extend your understanding:

1. How does the trigonometric identity \( \sin x \cos x = \frac{1}{2} \sin(2x) \) help simplify integrals?
2. What is the general approach to solving integrals that involve absolute value functions?
3. Why is integration by parts a useful technique for products of functions?
4. How does the integral \( \int \ln x \, dx \) derive from the integration by parts method?
5. What are some applications of definite integrals in physics and engineering?

**Tip:** When faced with a product of two functions in an integral, consider using integration by parts!

Evaluate the following definite integrals:

1. \( \int_0^\pi x \sin x \cos x \, dx \)
2. \( \int_{-1}^{1} |x| e^x \, dx \)
3. \( \int_{\frac{1}{e}}^e |\ln x| \, dx \)

Learn how to evaluate definite integrals such as \( \int_0^\pi x \sin x \cos x \, dx \), \( \int_{-1}^{1} |x| e^x \, dx \), and \( \int_{\frac{1}{e}}^e |\ln x| \, dx \). This detailed solution includes techniques like integration by parts and handling absolute value and logarithmic functions, suitable for university-level calculus students.

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