This looks a lot like homework, but since an answer has already been posted, here
are alternatives that do not use a transformation to polar coordinates
and needing to integrate trigonometric functions.
Since $\frac{\mathrm d}{\mathrm dx} \exp(-x^2/2) = -x\exp(-x^2/2)$, it is possible
to compute
$$P\{X^2+Y^2 \leq 4\} = \int_{x=0}^2 x\exp(-x^2/2)\int_{y=0}^{\sqrt{4-x^2}}
y\exp(-y^2/2)\,\mathrm dy\,\mathrm dx$$
without resorting to a transformation to polar coordinates.
The inner integral should work out to something like
$1 - \exp(-(4-x^2)/2) = 1 - e^{-2}\exp(x^2/2)$ which will change the
integrand of the outer integral to something like $x\exp(-x^2/2)-e^{-2}x$
which can also be easily
integrated.
As yet another approach, note that $X$ and $Y$ are independent
Rayleigh random variables and so $X = \sqrt{Z_1^2+Z_2^2},~ Y = \sqrt{Z_3^2+Z_4^2}$ where $Z_1, Z_2, Z_3, Z_4$ are independent
standard normal random variables. Thus
$P\{X^2+Y^2 > 4\}$ is the probability that a $\chi^2$ random
variable $W = X^2+Y^2 = Z_1^2+Z_2^2+Z_3^2+Z_4^2$ exceeds $4$.