1. **State the problem:** We need to evaluate the integral $$\int_{-2\pi}^{2\pi} x \cos x \, dx$$. 2. **Recall the formula and rules:** To solve this integral, we use integration by parts, which states: $$\int u \, dv = uv - \int v \, du$$ where we choose parts of the integrand as $u$ and $dv$. 3. **Choose $u$ and $dv$:** Let $$u = x \implies du = dx$$ $$dv = \cos x \, dx \implies v = \sin x$$ 4. **Apply integration by parts:** $$\int x \cos x \, dx = x \sin x - \int \sin x \, dx$$ 5. **Integrate $\int \sin x \, dx$:** $$\int \sin x \, dx = -\cos x$$ 6. **Substitute back:** $$\int x \cos x \, dx = x \sin x + \cos x + C$$ 7. **Evaluate definite integral:** $$\int_{-2\pi}^{2\pi} x \cos x \, dx = \left[ x \sin x + \cos x \right]_{-2\pi}^{2\pi}$$ 8. **Calculate at upper limit $2\pi$:** $$2\pi \sin(2\pi) + \cos(2\pi) = 2\pi \cdot 0 + 1 = 1$$ 9. **Calculate at lower limit $-2\pi$:** $$-2\pi \sin(-2\pi) + \cos(-2\pi) = -2\pi \cdot 0 + 1 = 1$$ 10. **Subtract:** $$1 - 1 = 0$$ **Final answer:** $$\int_{-2\pi}^{2\pi} x \cos x \, dx = 0$$