1. **State the problem:** We want to solve the integral $$\int x (\sin^2(x) - \cos^2(x)) \, dx$$. 2. **Use trigonometric identities:** Recall that $$\sin^2(x) - \cos^2(x) = -\cos(2x)$$. 3. **Rewrite the integral:** Substitute the identity into the integral: $$\int x (\sin^2(x) - \cos^2(x)) \, dx = \int x (-\cos(2x)) \, dx = -\int x \cos(2x) \, dx$$. 4. **Use integration by parts:** Let $$u = x$$ and $$dv = \cos(2x) dx$$. Then, $$du = dx$$ and $$v = \frac{\sin(2x)}{2}$$. 5. **Apply integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ So, $$-\int x \cos(2x) \, dx = -\left(x \cdot \frac{\sin(2x)}{2} - \int \frac{\sin(2x)}{2} \, dx\right) = -\frac{x}{2} \sin(2x) + \frac{1}{2} \int \sin(2x) \, dx$$. 6. **Integrate $$\int \sin(2x) dx$$:** $$\int \sin(2x) \, dx = -\frac{\cos(2x)}{2} + C$$. 7. **Substitute back:** $$-\frac{x}{2} \sin(2x) + \frac{1}{2} \left(-\frac{\cos(2x)}{2}\right) + C = -\frac{x}{2} \sin(2x) - \frac{1}{4} \cos(2x) + C$$. **Final answer:** $$-\frac{x}{2} \sin(2x) - \frac{1}{4} \cos(2x) + C$$ which corresponds to option b).