calculus - Integral of gaussian error function

Problem : Integrate the function $x\cdot e^{-x^2(1+y^2)}$ over the set $(0,\infty)\times(0,\infty)$ in two different ways. Conclude from your calculations that $\int_{0}^{\infty} e^{-t^2}\,dt = \frac{1}{2}\sqrt{\pi}$.

My attempt : I am trying to use Fubini theorem to approach the problem, I have so far calculated the integral by first integrating over $x$ and then $y$ and determined its value to be $\pi/4$ but I don't see how can I calculate it any other way (without knowing the integral of the gaussian error function first or how can I relate it to the integral of gaussian error function).

The problem essentially requires to work with Fubini's theorem in both the directions (or at least that is what my intuition is).

Answer

$$\iint_{(0,+\infty)^2}x e^{-x^2(1+y^2)}\,dx\,dy = \frac{1}{2}\int_{0}^{+\infty}\frac{dy}{1+y^2}=\frac{\pi}{4}$$
and
$$\begin{eqnarray*}\iint_{(0,+\infty)^2}x e^{-x^2(1+y^2)}\,dy\,dx &=& \int_{0}^{+\infty}\int_{0}^{+\infty} x e^{-x^2} e^{-(xy)^2}\,dy\,dx\\ & \stackrel{y\mapsto \frac{z}{x}}{=}& \int_{0}^{+\infty}\int_{0}^{+\infty} e^{-x^2} e^{-z^2}\,dz\,dx \end{eqnarray*}$$
equals the square of $\int_{0}^{+\infty} e^{-u^2}\,du$, hence $\int_{0}^{+\infty} e^{-u^2}\,du = \frac{\sqrt{\pi}}{2}$.
No polar coordinates, no $\Gamma$ function, just elementary manipulations. I like this approach!