To integrate \( \sin^3(4x) \), we can use trigonometric identities to simplify the expression.

### Step-by-Step Solution

1. **Rewrite \( \sin^3(4x) \) using the identity**:
   \[
   \sin^3(4x) = \sin(4x) \cdot \sin^2(4x)
   \]
   
   Since \( \sin^2(4x) = 1 - \cos^2(4x) \), we can rewrite \( \sin^3(4x) \) as:
   \[
   \sin^3(4x) = \sin(4x) \cdot (1 - \cos^2(4x))
   \]

2. **Expand the expression**:
   \[
   \sin^3(4x) = \sin(4x) - \sin(4x) \cdot \cos^2(4x)
   \]

3. **Integrate each term separately**:
   Now, the integral becomes:
   \[
   \int \sin^3(4x) \, dx = \int \sin(4x) \, dx - \int \sin(4x) \cdot \cos^2(4x) \, dx
   \]

4. **Solve the first integral**:
   \[
   \int \sin(4x) \, dx = -\frac{1}{4} \cos(4x)
   \]

5. **Use substitution for the second integral**:
   Let \( u = \cos(4x) \), so \( du = -4 \sin(4x) \, dx \) or \( \sin(4x) \, dx = -\frac{1}{4} \, du \).
   
   Substituting into the second integral, we get:
   \[
   \int \sin(4x) \cdot \cos^2(4x) \, dx = -\frac{1}{4} \int u^2 \, du
   \]

6. **Integrate \( u^2 \)**:
   \[
   -\frac{1}{4} \int u^2 \, du = -\frac{1}{4} \cdot \frac{u^3}{3} = -\frac{u^3}{12}
   \]

7. **Substitute back \( u = \cos(4x) \)**:
   \[
   -\frac{u^3}{12} = -\frac{\cos^3(4x)}{12}
   \]

8. **Combine results**:
   \[
   \int \sin^3(4x) \, dx = -\frac{1}{4} \cos(4x) + \frac{\cos^3(4x)}{12} + C
   \]

### Final Answer
\[
\int \sin^3(4x) \, dx = -\frac{1}{4} \cos(4x) + \frac{\cos^3(4x)}{12} + C
\]

Would you like more details or have any questions about this solution?

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Here are five questions to further explore this topic:

1. How would you integrate \( \cos^3(4x) \)?
2. Can you evaluate \( \int \sin^4(4x) \, dx \)?
3. How would the solution change if we integrated \( \sin^3(ax) \) with a general coefficient \( a \)?
4. What other methods could simplify trigonometric integrals like \( \sin^3(4x) \)?
5. How would you approach an integral involving \( \sin^3(4x) \cos^3(4x) \)?

**Tip**: Using substitution and trigonometric identities can make complex integrals much simpler by reducing the power of trigonometric functions.

Learn how to integrate sin^3(4x) using trigonometric identities and substitution. This guide provides a detailed solution, breaking down complex trigonometric integrals. Ideal for high school students in Grades 11-12.

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