1. **State the problem:** We need to evaluate the double integral $$\int_{-2}^2 \int_x^2 \cos(\sqrt{y^3}) \, dy \, dx.$$\n\n2. **Understand the integral:** The outer integral is with respect to $x$ from $-2$ to $2$, and the inner integral is with respect to $y$ from $y=x$ to $y=2$.\n\n3. **Consider the region of integration:** The region is defined by $-2 \leq x \leq 2$ and $x \leq y \leq 2$. This is a triangular region above the line $y=x$ and below $y=2$.\n\n4. **Change the order of integration:** It is often easier to integrate by changing the order. For $y$, the limits are from $-2$ to $2$ (since $y$ goes from the lowest $x$ to 2). For each fixed $y$, $x$ goes from $-2$ up to $y$ (since $x \leq y$). So the integral becomes:\n$$\int_{y=-2}^2 \int_{x=-2}^y \cos(\sqrt{y^3}) \, dx \, dy.$$\n\n5. **Integrate with respect to $x$ first:** Since the integrand $\cos(\sqrt{y^3})$ does not depend on $x$, the inner integral is:\n$$\int_{x=-2}^y \cos(\sqrt{y^3}) \, dx = \cos(\sqrt{y^3}) (y - (-2)) = (y+2) \cos(\sqrt{y^3}).$$\n\n6. **Rewrite the integral:** Now the integral reduces to a single integral:\n$$\int_{-2}^2 (y+2) \cos(\sqrt{y^3}) \, dy.$$\n\n7. **Evaluate the integral:** This integral is complicated due to the $\cos(\sqrt{y^3})$ term. There is no elementary antiderivative. However, since the problem asks for the whole integral, we can consider symmetry or numerical approximation.\n\n8. **Check symmetry:** The function inside is $f(y) = (y+2) \cos(\sqrt{y^3})$. Note that $\sqrt{y^3}$ is not defined for negative $y$ in the real numbers because $y^3$ is negative and square root of negative is complex. So the integral is not real-valued over negative $y$.\n\n9. **Domain consideration:** Since $\sqrt{y^3}$ is real only for $y \geq 0$, the integral over $y$ from $-2$ to $0$ is not real. Therefore, the integral as stated is not defined over the entire domain.\n\n10. **Conclusion:** The integral $$\int_{-2}^2 \int_x^2 \cos(\sqrt{y^3}) \, dy \, dx$$ is not defined in the real numbers because $\sqrt{y^3}$ is not real for negative $y$.\n\n**Final answer:** The integral is not defined over the given domain in the real numbers due to the square root of negative values inside the cosine function.