Square roots in integrals generally make problems difficult, but here it's a clue that the problem can be simplified by making a change of variable to polar coordinates. Let $x = r \cos \theta$, $y=r\sin\theta$.
To get you started, we write the region of integration in polar coordinates. The closed disk $x^2+y^2\leq 4$ is given in polar coordinates by $\{(r,\theta)|0\leq r\leq 2, 0 \leq \theta < 2\pi\}$. This gives you very easy limits of integration.
You've already found the integrand in terms of $x$ and $y$. Use the polar coordinate transformation to express it in terms of $r$ and $\theta$, and use trigonometric identities to make the integrand very simple.
Finally, remember that whenever you make a change in variable you must multiply by the appropriate Jacobian determinant. (It's easy to forget this step. Remember that $dA=dxdy$ in our original coordinates, but $dA\neq drd\theta$ in our new coordinates - something's missing.)