To solve the integral $\int x \sin(2x) \, dx$ using integration by parts, we follow these steps:

Step 1: Choose $u$ and $dv$

Integration by parts is based on the formula:

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

We choose:

• $u = x$ (so that its derivative simplifies)
• $dv = \sin(2x) \, dx$ (so we can integrate it)

Step 2: Differentiate $u$ and Integrate $dv$

Now, we compute $du$ and $v$:

• $du = dx$ (since the derivative of $x$ is 1)
• To find $v$, we integrate $dv = \sin(2x) \, dx$: $v = \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x)$ (using the standard result for integrating $\sin(kx)$).

Step 3: Apply the Integration by Parts Formula

Now we apply the integration by parts formula:

$\int x \sin(2x) \, dx = u v - \int v \, du$ Substitute $u = x$, $v = -\frac{1}{2} \cos(2x)$, and $du = dx$:

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

Step 4: Integrate the Remaining Integral

Now, integrate $\int \cos(2x) \, dx$:

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

Step 5: Final Answer

Substitute this result back into the expression:

$-\frac{1}{2} x \cos(2x) + \frac{1}{2} \left(\frac{1}{2} \sin(2x)\right) + C$ $= -\frac{1}{2} x \cos(2x) + \frac{1}{4} \sin(2x) + C$

Thus, the final solution is:

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

Let me know if you'd like further clarification or have additional questions!

Here are five related questions for further practice:

1. What is the integration by parts formula and how does it work?
2. Can you solve $\int x e^{2x} \, dx$ using integration by parts?
3. How do you apply integration by parts when multiple iterations are needed?
4. Can integration by parts be used to evaluate definite integrals?
5. How do you choose $u$ and $dv$ efficiently for integration by parts?

Tip: Integration by parts is often easier if you choose $u$ such that its derivative simplifies the expression (e.g., powers of $x$).