1. **Problem statement:** Evaluate the contour integral $$\int_C \frac{\cos x}{x^2 + 1} \, dx$$ where $C$ is the circle $|x|=2$, using Cauchy's residue theorem. 2. **Recall Cauchy's residue theorem:** For a function $f$ analytic inside and on a simple closed contour $C$ except for isolated singularities inside $C$, the integral is $$\int_C f(z) \, dz = 2\pi i \sum \text{Residues of } f \text{ at singularities inside } C.$$ 3. **Identify singularities:** The denominator $x^2 + 1 = 0$ gives singularities at $x = i$ and $x = -i$. Both lie inside the circle $|x|=2$ since $|i|=1 < 2$ and $|-i|=1 < 2$. 4. **Calculate residues:** The function is $$f(x) = \frac{\cos x}{x^2 + 1} = \frac{\cos x}{(x - i)(x + i)}.$$ The singularities are simple poles. Residue at $x = i$: $$\text{Res}(f, i) = \lim_{x \to i} (x - i) \frac{\cos x}{(x - i)(x + i)} = \lim_{x \to i} \frac{\cos x}{x + i} = \frac{\cos i}{i + i} = \frac{\cos i}{2i}.$$ Residue at $x = -i$: $$\text{Res}(f, -i) = \lim_{x \to -i} (x + i) \frac{\cos x}{(x - i)(x + i)} = \lim_{x \to -i} \frac{\cos x}{x - i} = \frac{\cos(-i)}{-i - i} = \frac{\cos(-i)}{-2i}.$$ Since $\cos$ is even, $\cos(-i) = \cos i$. 5. **Sum of residues:** $$\text{Res}(f, i) + \text{Res}(f, -i) = \frac{\cos i}{2i} + \frac{\cos i}{-2i} = \frac{\cos i}{2i} - \frac{\cos i}{2i} = 0.$$ 6. **Conclusion:** The sum of residues inside $C$ is zero, so by Cauchy's residue theorem, $$\int_C \frac{\cos x}{x^2 + 1} \, dx = 2\pi i \times 0 = 0.$$ **Final answer:** $$\boxed{0}.$$