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