1. **State the problem:** Evaluate the integral $$\int (\sin 3x \cos x - \cos 3x \sin x) \, dx$$. 2. **Recognize the trigonometric identity:** The expression inside the integral matches the sine difference formula: $$\sin A \cos B - \cos A \sin B = \sin(A - B)$$ Here, $A = 3x$ and $B = x$, so: $$\sin 3x \cos x - \cos 3x \sin x = \sin(3x - x) = \sin 2x$$ 3. **Rewrite the integral:** $$\int (\sin 3x \cos x - \cos 3x \sin x) \, dx = \int \sin 2x \, dx$$ 4. **Integrate:** Recall that: $$\int \sin kx \, dx = -\frac{1}{k} \cos kx + C$$ So, $$\int \sin 2x \, dx = -\frac{1}{2} \cos 2x + C$$ 5. **Final answer:** $$-\frac{1}{2} \cos 2x + C$$ This corresponds to option (b).