Integral xcos(8x)dx; u = x, dv=cos(8x) dx evaluate the integral using integral by parts

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.

Learn how to evaluate the integral ∫xcos(8x) dx using integration by parts. This solution includes step-by-step explanations, applying the integration by parts formula and solving trigonometric integrals. Suitable for advanced high school students and undergraduate calculus courses.

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