1. The problem is to evaluate the integral $$\int \cos(4x) \sin(2x) \, dx$$. 2. Use the product-to-sum identity for sine and cosine: $$\cos A \sin B = \frac{1}{2} [\sin(A+B) - \sin(A-B)]$$. 3. Substitute $A=4x$ and $B=2x$: $$\cos(4x) \sin(2x) = \frac{1}{2} [\sin(4x+2x) - \sin(4x-2x)] = \frac{1}{2} [\sin(6x) - \sin(2x)]$$. 4. Rewrite the integral: $$\int \cos(4x) \sin(2x) \, dx = \int \frac{1}{2} [\sin(6x) - \sin(2x)] \, dx = \frac{1}{2} \int \sin(6x) \, dx - \frac{1}{2} \int \sin(2x) \, dx$$. 5. Integrate each term: - $$\int \sin(6x) \, dx = -\frac{1}{6} \cos(6x) + C$$ - $$\int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C$$ 6. Substitute back: $$\frac{1}{2} \left(-\frac{1}{6} \cos(6x)\right) - \frac{1}{2} \left(-\frac{1}{2} \cos(2x)\right) + C = -\frac{1}{12} \cos(6x) + \frac{1}{4} \cos(2x) + C$$. 7. Final answer: $$\int \cos(4x) \sin(2x) \, dx = -\frac{1}{12} \cos(6x) + \frac{1}{4} \cos(2x) + C$$.