To integrate \( \cos^3(3x) \, dx \), we can use a trigonometric identity and substitution. Here are the steps:

### Step 1: Rewrite \( \cos^3(3x) \)
We can use the identity:
\[
\cos^3(3x) = \cos^2(3x) \cdot \cos(3x) = (1 - \sin^2(3x)) \cos(3x)
\]

Thus,
\[
\int \cos^3(3x) \, dx = \int (1 - \sin^2(3x)) \cos(3x) \, dx
\]

### Step 2: Substitute \( u = \sin(3x) \)
Let \( u = \sin(3x) \). Then \( \frac{du}{dx} = 3 \cos(3x) \), or \( dx = \frac{du}{3 \cos(3x)} \).

Substituting, we get:
\[
\int (1 - u^2) \cos(3x) \, \frac{du}{3 \cos(3x)} = \frac{1}{3} \int (1 - u^2) \, du
\]

### Step 3: Integrate with respect to \( u \)
Now we have:
\[
\frac{1}{3} \int (1 - u^2) \, du = \frac{1}{3} \left( u - \frac{u^3}{3} \right) + C
\]

### Step 4: Substitute back \( u = \sin(3x) \)
Replacing \( u \) with \( \sin(3x) \), we get:
\[
\frac{1}{3} \left( \sin(3x) - \frac{\sin^3(3x)}{3} \right) + C
\]

### Final Answer
\[
\int \cos^3(3x) \, dx = \frac{1}{3} \sin(3x) - \frac{1}{9} \sin^3(3x) + C
\]

Would you like further clarification on any part of this process?

### Related Questions:
1. How would you integrate \( \cos^3(x) \) without the \(3x\) factor?
2. Can this method be applied to \( \cos^n(kx) \) for other powers \( n \)?
3. How would you handle \( \sin^3(3x) \) instead of \( \cos^3(3x) \)?
4. What are other techniques for integrating trigonometric functions raised to a power?
5. How would you integrate \( \cos^4(3x) \) or \( \cos^5(3x) \)?

### Tip:
When integrating powers of trigonometric functions, rewriting the function using trigonometric identities often simplifies the integral, making substitution possible.

Learn how to integrate cos^3(3x) using trigonometric identities and substitution. This detailed solution includes step-by-step instructions and shows how to simplify the integral, making it easy to solve. Ideal for high school students learning advanced integration techniques.

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