Answer is: option3

\(-\frac{x}{2} \cos(2x) + \frac{1}{4} \sin(2x) + C\)

Solution:

We are tasked with finding the value of the following integral:

\[ \int x \sin(2x) \, dx \]

Step 1: Use Integration by Parts

We will use the integration by parts formula:

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

Let us set:

\[ u = x \quad \Rightarrow \quad du = dx \]

\[ dv = \sin(2x) \, dx \quad \Rightarrow \quad v = -\frac{1}{2} \cos(2x) \]

Applying integration by parts to \( \int x \sin(2x) \, dx \), we get:

\[ \int x \sin(2x) \, dx = x \left( -\frac{1}{2} \cos(2x) \right) - \int \left( -\frac{1}{2} \cos(2x) \right) \, dx \]

Step 2: Solve the Remaining Integral

Now, simplify the equation:

\[ \int x \sin(2x) \, dx = -\frac{x}{2} \cos(2x) + \frac{1}{2} \int \cos(2x) \, dx \]

We know that:

\[ \int \cos(2x) \, dx = \frac{1}{2} \sin(2x) \]

Substitute this result into the equation:

\[ = -\frac{x}{2} \cos(2x) + \frac{1}{4} \sin(2x) \]

Step 3: Final Answer

Adding the constant of integration, we get:

\[ \int x \sin(2x) \, dx = -\frac{x}{2} \cos(2x) + \frac{1}{4} \sin(2x) + C \]