To integrate $\sin^3(3x)$, we use trigonometric identities and substitution.

Steps:

1. Rewrite $\sin^3(3x)$: Using the identity $\sin^3(x) = \sin(x)(1 - \cos^2(x))$: $\sin^3(3x) = \sin(3x)(1 - \cos^2(3x)).$

2. Separate into two terms: $\int \sin^3(3x) \, dx = \int \sin(3x) \, dx - \int \sin(3x)\cos^2(3x) \, dx.$

3. First integral ($\int \sin(3x) \, dx$): The integral of $\sin(3x)$ is straightforward: $\int \sin(3x) \, dx = -\frac{1}{3} \cos(3x).$

4. Second integral ($\int \sin(3x)\cos^2(3x) \, dx$): Use substitution. Let $u = \cos(3x)$, so $du = -3\sin(3x) \, dx$. This transforms the integral: $\int \sin(3x)\cos^2(3x) \, dx = -\frac{1}{3} \int u^2 \, du.$ Compute the integral of $u^2$: $-\frac{1}{3} \int u^2 \, du = -\frac{1}{3} \cdot \frac{u^3}{3} = -\frac{u^3}{9}.$ Substituting back $u = \cos(3x)$: $-\frac{u^3}{9} = -\frac{\cos^3(3x)}{9}.$

5. Combine results: $\int \sin^3(3x) \, dx = -\frac{1}{3} \cos(3x) - \frac{\cos^3(3x)}{9} + C.$

Final Answer:

$\int \sin^3(3x) \, dx = -\frac{1}{3} \cos(3x) - \frac{\cos^3(3x)}{9} + C.$

Would you like a deeper explanation or details on any of these steps?

---

Relative Questions:

1. How does the substitution method work in integration?
2. Can you explain the trigonometric identity $\sin^3(x) = \sin(x)(1 - \cos^2(x))$?
3. What are common strategies for integrating powers of sine and cosine?
4. How do you verify the correctness of an integral result?
5. How would this integration change if the exponent were 4 instead of 3?

Tip: When integrating powers of trigonometric functions, breaking them into simpler components often makes the problem more manageable.