To solve the integral of $\sin^3(3x)$, we can use a reduction formula and trigonometric identities. Here's the step-by-step process:

Step 1: Express $\sin^3(3x)$ using trigonometric identities

We use the identity: $\sin^3(\theta) = \sin(\theta) \cdot (1 - \cos^2(\theta))$ So, $\sin^3(3x) = \sin(3x) \cdot (1 - \cos^2(3x))$

Step 2: Break the integral into two parts

Now the integral becomes: $\int \sin^3(3x) \, dx = \int \sin(3x) \, dx - \int \sin(3x) \cos^2(3x) \, dx$

Step 3: Solve the first integral

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

Step 4: Solve the second integral using substitution

For the second integral, let’s use substitution. Let: $u = \cos(3x) \quad \text{so} \quad du = -3 \sin(3x) \, dx \quad \text{or} \quad dx = \frac{du}{-3 \sin(3x)}$ This transforms the second integral into: $-\int \sin(3x) \cos^2(3x) \, dx = -\frac{1}{3} \int u^2 \, du$

Step 5: Integrate $u^2$

The integral of $u^2$ is: $\int u^2 \, du = \frac{u^3}{3} = \frac{\cos^3(3x)}{3}$

Step 6: Combine the results

Now, combining both parts, we have: $\int \sin^3(3x) \, dx = -\frac{1}{3} \cos(3x) + \frac{1}{9} \cos^3(3x) + C$

Thus, the integral of $\sin^3(3x)$ is: $\boxed{-\frac{1}{3} \cos(3x) + \frac{1}{9} \cos^3(3x) + C}$

Would you like more details or have any questions?

Here are 5 related questions:

1. How do you integrate powers of sine and cosine in general?
2. What are reduction formulas for trigonometric integrals?
3. How does substitution work in trigonometric integrals?
4. How can you express higher powers of trigonometric functions?
5. What is the importance of trigonometric identities in integration?

Tip: When dealing with powers of sine or cosine, consider using trigonometric identities to simplify the integrand first.