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.