1. **State the problem:** Evaluate the integral $\int \cos(2x - 4x) \, dx$. 2. **Simplify the argument of cosine:** $$2x - 4x = -2x$$ So the integral becomes: $$\int \cos(-2x) \, dx$$ 3. **Use the even property of cosine:** Since $\cos(-\theta) = \cos(\theta)$, we have: $$\int \cos(-2x) \, dx = \int \cos(2x) \, dx$$ 4. **Recall the integral formula:** $$\int \cos(ax) \, dx = \frac{1}{a} \sin(ax) + C$$ where $a$ is a constant. 5. **Apply the formula:** Here, $a = 2$, so: $$\int \cos(2x) \, dx = \frac{1}{2} \sin(2x) + C$$ 6. **Final answer:** $$\boxed{\frac{1}{2} \sin(2x) + C}$$