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.