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The integral you have uploaded is:

\[
\int x \sin(x) \, dx
\]

To solve this, we use **integration by parts**, which is based on the formula:

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

For this problem, let's choose:

- \( u = x \), so \( du = dx \)
- \( dv = \sin(x) \, dx \), so \( v = -\cos(x) \)

Now apply the integration by parts formula:

\[
\int x \sin(x) \, dx = -x \cos(x) + \int \cos(x) \, dx
\]

The remaining integral is straightforward:

\[
\int \cos(x) \, dx = \sin(x)
\]

Thus, the complete solution is:

\[
\int x \sin(x) \, dx = -x \cos(x) + \sin(x) + C
\]

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

Would you like further details or have any questions? 

Here are 5 related questions to explore further:
1. How does integration by parts work in general cases?
2. What are some common integrals involving trigonometric functions?
3. How would you apply integration by parts to \(\int x \cos(x) dx\)?
4. Can you use a different method to solve \(\int x \sin(x) dx\)?
5. How does integration by parts relate to the product rule in differentiation?

**Tip:** When applying integration by parts, choosing \( u \) and \( dv \) wisely (where \( dv \) is easier to integrate) can simplify the process.

Learn how to solve the integral ∫ x sin(x) dx using the integration by parts method. This step-by-step solution covers key calculus concepts, including integration by parts and trigonometric functions, ideal for college-level students.

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