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The given problem asks to compute the definite integral:

\[
I = \int_0^\pi x^2 \cos(x) \, dx
\]

This problem suggests using **integration by parts**. Integration by parts is based on the formula:

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

### Step 1: Choose \( u \) and \( dv \)
We can choose:
- \( u = x^2 \), which means \( du = 2x \, dx \)
- \( dv = \cos(x) \, dx \), which means \( v = \sin(x) \)

### Step 2: Apply the integration by parts formula
Substitute into the formula:

\[
\int_0^\pi x^2 \cos(x) \, dx = \left[ x^2 \sin(x) \right]_0^\pi - \int_0^\pi 2x \sin(x) \, dx
\]

### Step 3: Evaluate the boundary term
Now, evaluate the first part \( \left[ x^2 \sin(x) \right]_0^\pi \):
- At \( x = \pi \), \( \sin(\pi) = 0 \), so \( \pi^2 \sin(\pi) = 0 \).
- At \( x = 0 \), \( \sin(0) = 0 \), so \( 0^2 \sin(0) = 0 \).

Thus, the boundary term is zero:

\[
\left[ x^2 \sin(x) \right]_0^\pi = 0
\]

Now we are left with the integral:

\[
I = - \int_0^\pi 2x \sin(x) \, dx
\]

### Step 4: Apply integration by parts again
For this integral, we apply integration by parts again:
- Let \( u = 2x \), so \( du = 2 \, dx \)
- Let \( dv = \sin(x) \, dx \), so \( v = -\cos(x) \)

Now apply the formula:

\[
- \int_0^\pi 2x \sin(x) \, dx = - \left[ -2x \cos(x) \right]_0^\pi + \int_0^\pi 2 \cos(x) \, dx
\]

### Step 5: Evaluate the boundary term
Evaluate \( \left[ -2x \cos(x) \right]_0^\pi \):
- At \( x = \pi \), \( \cos(\pi) = -1 \), so \( -2\pi \cos(\pi) = 2\pi \).
- At \( x = 0 \), \( \cos(0) = 1 \), so \( -2(0) \cos(0) = 0 \).

Thus, the boundary term evaluates to:

\[
- \left[ -2x \cos(x) \right]_0^\pi = 2\pi
\]

Now we are left with:

\[
2\pi + \int_0^\pi 2 \cos(x) \, dx
\]

### Step 6: Evaluate the remaining integral
The remaining integral is straightforward:

\[
\int_0^\pi 2 \cos(x) \, dx = 2 \sin(x) \bigg|_0^\pi
\]

At \( x = \pi \), \( \sin(\pi) = 0 \), and at \( x = 0 \), \( \sin(0) = 0 \), so:

\[
\int_0^\pi 2 \cos(x) \, dx = 0
\]

### Step 7: Final answer
Thus, the final result of the integral is:

\[
I = 2\pi
\]

The correct answer is:

\[
\boxed{2\pi}
\]

Let me know if you'd like more details or further clarification.

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Here are some related questions you might want to explore:
1. How is integration by parts derived from the product rule?
2. Can you apply integration by parts to an integral like \( \int x^3 e^x \, dx \)?
3. How would you solve an integral involving trigonometric identities?
4. What are the advantages of using definite integrals over indefinite ones?
5. How does the result change if we change the limits of integration in this problem?

**Tip:** When dealing with complex integrals, it helps to break them down into simpler steps like applying integration by parts or recognizing common patterns.

Calculate the integral: \( \int_0^\pi x^2 \cos(x) \, dx \)

This problem demonstrates the use of integration by parts to solve the definite integral \( \int_0^\pi x^2 \cos(x) \, dx \). Learn the step-by-step breakdown, including applying trigonometric identities and evaluating definite integrals. Suitable for university-level calculus students.

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