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.

Learn how to solve the integral of sin^3(3x) step by step using trigonometric identities and the substitution method. This detailed explanation includes reduction formulas and substitution techniques, making it suitable for college-level calculus students.

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