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.