1. **State the problem:** We need to solve the integral $$\int 4x \cos^2(x) \, dx$$. 2. **Use the double-angle identity:** Recall that $$\cos^2(x) = \frac{1 + \cos(2x)}{2}$$. 3. **Rewrite the integral:** $$\int 4x \cos^2(x) \, dx = \int 4x \cdot \frac{1 + \cos(2x)}{2} \, dx = \int 2x (1 + \cos(2x)) \, dx = \int 2x \, dx + \int 2x \cos(2x) \, dx$$. 4. **Integrate the first part:** $$\int 2x \, dx = x^2 + C_1$$. 5. **Integrate the second part using integration by parts:** Let $$u = 2x$$ and $$dv = \cos(2x) dx$$. Then $$du = 2 dx$$ and $$v = \frac{\sin(2x)}{2}$$. Using integration by parts formula $$\int u \, dv = uv - \int v \, du$$: $$\int 2x \cos(2x) \, dx = 2x \cdot \frac{\sin(2x)}{2} - \int \frac{\sin(2x)}{2} \cdot 2 \, dx = x \sin(2x) - \int \sin(2x) \, dx$$. 6. **Integrate $$\int \sin(2x) dx$$:** $$\int \sin(2x) \, dx = -\frac{\cos(2x)}{2} + C_2$$. 7. **Combine results:** $$\int 2x \cos(2x) \, dx = x \sin(2x) + \frac{\cos(2x)}{2} + C_3$$. 8. **Final integral:** $$\int 4x \cos^2(x) \, dx = x^2 + x \sin(2x) + \frac{\cos(2x)}{2} + C$$. 9. **Match with options:** This corresponds to option (a): $$x^2 + x \sin(2x) + \frac{1}{2} \cos(2x) + C$$.