To integrate \( \sin^3(4x) \), we can use the identity \( \sin^3(x) = \sin(x) \cdot (1 - \cos^2(x)) \) to simplify the expression:

1. Rewrite \( \sin^3(4x) \):
   \[
   \sin^3(4x) = \sin(4x) \cdot (1 - \cos^2(4x))
   \]
   
2. This gives us:
   \[
   \int \sin^3(4x) \, dx = \int \sin(4x) \cdot (1 - \cos^2(4x)) \, dx
   \]

3. Use substitution. Let \( u = \cos(4x) \), so \( \frac{du}{dx} = -4 \sin(4x) \), or \( \sin(4x) \, dx = -\frac{du}{4} \).

4. Substitute into the integral:
   \[
   = -\frac{1}{4} \int (1 - u^2) \, du
   \]

5. Expand and integrate each term:
   \[
   = -\frac{1}{4} \int (1 - u^2) \, du = -\frac{1}{4} \left( u - \frac{u^3}{3} \right) + C
   \]

6. Substitute back \( u = \cos(4x) \):
   \[
   = -\frac{1}{4} \left( \cos(4x) - \frac{\cos^3(4x)}{3} \right) + C
   \]

Therefore, the result is:
\[
\int \sin^3(4x) \, dx = -\frac{1}{4} \cos(4x) + \frac{1}{12} \cos^3(4x) + C
\]

Would you like more details on the substitution process or have any questions?

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**Follow-up Questions:**
1. How would you integrate \( \sin^n(kx) \) for other odd powers of \( n \)?
2. What identity would be useful if we had \( \cos^3(4x) \) instead?
3. Can you explain why substitution was chosen here?
4. What other methods could solve integrals with powers of trigonometric functions?
5. How does changing the coefficient inside sine, e.g., \( \sin^3(5x) \), affect the integration?

**Tip:** When working with powers of sine or cosine, splitting off one sine or cosine factor can often simplify integration.

Explore the step-by-step integration of sin^3(4x) by applying trigonometric identities and the substitution method. This problem covers advanced trigonometric integration techniques suitable for Grades 11-12 students.

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