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.