To solve the integral $\int x \cos(8x) \, dx$ using integration by parts, we can follow the formula for integration by parts:

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

In this case, we are given:

• $u = x$
• $dv = \cos(8x) \, dx$

Step 1: Differentiate $u$ and integrate $dv$

• $du = dx$ (since the derivative of $x$ is 1)

• Integrating $dv = \cos(8x) \, dx$ gives:

  $v = \frac{1}{8} \sin(8x)$

Step 2: Apply the formula for integration by parts

Now apply the formula:

$\int x \cos(8x) \, dx = uv - \int v \, du$

Substitute $u = x$, $v = \frac{1}{8} \sin(8x)$, and $du = dx$:

$\int x \cos(8x) \, dx = x \cdot \frac{1}{8} \sin(8x) - \int \frac{1}{8} \sin(8x) \, dx$

Step 3: Evaluate the remaining integral

The remaining integral is:

$\int \sin(8x) \, dx = -\frac{1}{8} \cos(8x)$

So, we have:

$\int x \cos(8x) \, dx = \frac{1}{8} x \sin(8x) + \frac{1}{64} \cos(8x) + C$

where $C$ is the constant of integration.

Final Answer:

$\int x \cos(8x) \, dx = \frac{1}{8} x \sin(8x) + \frac{1}{64} \cos(8x) + C$

Would you like any more details or explanations?

Here are 5 related questions:

1. How does integration by parts simplify integrals involving products?
2. What happens when you reverse the roles of $u$ and $dv$ in integration by parts?
3. Can you apply integration by parts multiple times? If yes, when is this necessary?
4. How would the solution change if the integral were $\int x \sin(8x) \, dx$?
5. How do definite integrals change the process of integration by parts?

Tip: Integration by parts is particularly useful for products of polynomial and trigonometric or exponential functions.