1. **State the problem:** We want to evaluate the integral $$I = \int \sin^3(4x) \cos^2(4x) \, dx.$$\n\n2. **Rewrite the integrand:** Use the identity $$\sin^3(4x) = \sin(4x) \cdot \sin^2(4x)$$ and recall that $$\sin^2(\theta) = 1 - \cos^2(\theta).$$ So,\n$$\sin^3(4x) = \sin(4x)(1 - \cos^2(4x)).$$\nThus, the integral becomes\n$$I = \int \sin(4x)(1 - \cos^2(4x)) \cos^2(4x) \, dx = \int \sin(4x)(\cos^2(4x) - \cos^4(4x)) \, dx.$$\n\n3. **Substitution:** Let $$u = \cos(4x).$$ Then, $$\frac{du}{dx} = -4 \sin(4x) \Rightarrow \sin(4x) dx = -\frac{1}{4} du.$$\n\n4. **Rewrite the integral in terms of $u$:**\n$$I = \int \sin(4x)(u^2 - u^4) \, dx = \int (u^2 - u^4) \sin(4x) dx = \int (u^2 - u^4) \left(-\frac{1}{4} du\right) = -\frac{1}{4} \int (u^2 - u^4) du.$$\n\n5. **Integrate:**\n$$-\frac{1}{4} \int (u^2 - u^4) du = -\frac{1}{4} \left( \frac{u^3}{3} - \frac{u^5}{5} \right) + C = -\frac{1}{4} \left( \frac{\cos^3(4x)}{3} - \frac{\cos^5(4x)}{5} \right) + C.$$\n\n6. **Simplify the final answer:**\n$$I = -\frac{1}{12} \cos^3(4x) + \frac{1}{20} \cos^5(4x) + C.$$\n\nThis is the evaluated integral of the given function.