We are tasked with evaluating the indefinite integral:

$\int x \sin(2x) \, dx$

The problem suggests using integration by parts. The general formula for integration by parts is:

$\int u \, dv = uv - \int v \, du$

Given:

• $u = x$
• $dv = \sin(2x) \, dx$

Step-by-Step Solution:

1. Differentiate $u$: $u = x \quad \Rightarrow \quad du = dx$

2. Integrate $dv$: $dv = \sin(2x) \, dx \quad \Rightarrow \quad v = \int \sin(2x) \, dx$ To integrate $\sin(2x)$, we use a simple substitution or recall the known integral: $\int \sin(2x) \, dx = -\frac{1}{2} \cos(2x)$

3. Apply the integration by parts formula: $\int x \sin(2x) \, dx = uv - \int v \, du$ Substituting: $\int x \sin(2x) \, dx = x \left( -\frac{1}{2} \cos(2x) \right) - \int \left( -\frac{1}{2} \cos(2x) \right) \, dx$ Simplifying: $= -\frac{1}{2} x \cos(2x) + \frac{1}{2} \int \cos(2x) \, dx$

4. Solve the remaining integral: $\int \cos(2x) \, dx = \frac{1}{2} \sin(2x)$ So the full expression becomes: $\int x \sin(2x) \, dx = -\frac{1}{2} x \cos(2x) + \frac{1}{4} \sin(2x) + C$

Now, let's analyze the given options:

Looking at the provided choices, the correct form should match this structure before integrating the second term. The correct answer is:

$-\frac{1}{2} x \cos(2x) + \frac{1}{2} \int \cos(2x) \, dx$

This matches the fourth option:

$\boxed{-\frac{1}{2} x \cos(2x) + \frac{1}{2} \int \cos(2x) \, dx}$

Would you like further details or explanations on any step? Here are some additional related questions for practice:

1. What is the integral of $\sin(3x)$?
2. How would you apply integration by parts to $\int x e^{2x} \, dx$?
3. What is the result of $\int x^2 \sin(2x) \, dx$?
4. How does the integration by parts formula change when integrating functions with powers of $x$?
5. How do you integrate $\int x^n e^{kx} \, dx$?

Tip: When applying integration by parts, always ensure you choose $u$ and $dv$ such that $du$ simplifies and $v$ is easy to integrate.