To solve the integral

$\int \sin^3(3x) \cos(3x) \, dx$

we proceed as follows:

Step 1: Simplify using substitution

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

Substituting these into the integral: $\int \sin^3(3x) \cos(3x) \, dx = \int u^3 \cdot \frac{1}{3} \, du$

This simplifies to: $\frac{1}{3} \int u^3 \, du$

Step 2: Integrate

The integral of $u^3$ is: $\int u^3 \, du = \frac{u^4}{4}$

So the result becomes: $\frac{1}{3} \cdot \frac{u^4}{4} = \frac{u^4}{12}$

Step 3: Back-substitute

Substituting back $u = \sin(3x)$, we get: $\frac{u^4}{12} = \frac{\sin^4(3x)}{12}$

Step 4: Add the constant of integration

Finally, the solution is: $\frac{\sin^4(3x)}{12} + C$

Final Answer:

$\int \sin^3(3x) \cos(3x) \, dx = \frac{\sin^4(3x)}{12} + C$

Would you like further clarification or detailed steps about substitution? Here are 5 related questions:

1. How does substitution work for integrals involving trigonometric powers?
2. Can you show examples of similar integrals using other trigonometric identities?
3. How do we verify the solution using differentiation?
4. What happens if we rewrite $\sin^3(3x)$ in terms of $\sin(3x)$ and $\cos(3x)$?
5. What are common techniques to simplify integrals with odd powers of sine or cosine?

Tip: For integrals with odd powers of sine or cosine, substitution often simplifies the process dramatically!