1. **State the problem:** We need to evaluate the integral $$I = \int 2 \cos^2 x \, dx$$ and determine which of the given options is correct. 2. **Recall the formula:** Use the double-angle identity for cosine squared: $$\cos^2 x = \frac{1 + \cos 2x}{2}$$ 3. **Rewrite the integral:** Substitute the identity into the integral: $$I = \int 2 \cos^2 x \, dx = \int 2 \cdot \frac{1 + \cos 2x}{2} \, dx = \int (1 + \cos 2x) \, dx$$ 4. **Integrate term-by-term:** $$I = \int 1 \, dx + \int \cos 2x \, dx = x + \int \cos 2x \, dx$$ 5. **Integrate $$\cos 2x$$:** Recall that $$\int \cos(ax) \, dx = \frac{\sin(ax)}{a} + C$$, so $$\int \cos 2x \, dx = \frac{\sin 2x}{2} + C$$ 6. **Combine results:** $$I = x + \frac{\sin 2x}{2} + C$$ 7. **Conclusion:** The correct integral expression is $$I = x + \frac{1}{2} \sin 2x + C$$ This matches the option: "I = x + 1/2 sin 2x + C". **Final answer:** $$I = x + \frac{1}{2} \sin 2x + C$$