To integrate $\sin^5(3x) \cos(3x)$, we can use a substitution method and the fact that the sine and cosine functions are related. Here's how:

1. Substitute: Let $u = \sin(3x)$, so that $du = 3\cos(3x)dx$, or $\frac{du}{3} = \cos(3x)dx$.

2. Rewrite the integral: The integral becomes: $\int \sin^5(3x) \cos(3x) dx = \int u^5 \cdot \frac{du}{3}$ Simplify the constants: $\frac{1}{3} \int u^5 du$

3. Integrate: Now, integrate $u^5$: $\frac{1}{3} \cdot \frac{u^6}{6} = \frac{u^6}{18}$

4. Substitute back $u = \sin(3x)$: $\frac{(\sin(3x))^6}{18} + C$

Thus, the integral is: $\int \sin^5(3x) \cos(3x) dx = \frac{\sin^6(3x)}{18} + C$

Do you want more details or have any questions?

Here are some related questions:

1. How would you integrate $\sin^3(3x) \cos(3x)$?
2. What happens when you integrate $\cos^5(3x) \sin(3x)$?
3. Can you solve the integral $\sin^7(3x) \cos(3x) dx$?
4. How would you apply substitution to solve $\sin(3x) \cos^5(3x) dx$?
5. What is the integral of $\cos(3x) \sin^3(3x) dx$?

Tip: When you see products of sine and cosine powers, substitution and using trigonometric identities often simplifies the process.