$$\iiiint_{x^2+y^2+u^2+v^2\leq 1}e^{x^2+y^2-u^2-v^2}\,dx\,dy\,du\,dv$$
So we just learned substitution and i thought maybe for this integral doing 2 polar subs for x,y and for u,v but i'm not sure this is the right way for this.
Any hints will be welcome
Nothing wrong with what you said, we'll just have to do an additional substitution afterwards. Let
$$\begin{cases} x = r\cos\theta \\ y = r\sin\theta \\ u = s\cos \phi \\ v = s\sin\phi \\ \end{cases} \implies J = rs$$
which gives us the integral
$$\int_{r^2+s^2\leq 1 \:\cap\:(\theta,\phi)\in[0,2\pi]^2} rse^{r^2-s^2}\:dr\:ds\:d\theta\:d\phi = 4\pi^2\int_{r^2+s^2\leq 1} rs e^{r^2-s^2}\:dr\:ds$$
Now let
$$\begin{cases}r = \rho \cos\gamma \\ s = \rho \sin\gamma \\ \end{cases} \implies J = \rho$$
giving us the integral
$$4\pi^2 \int_0^1 \int_0^{\frac{\pi}{2}} \rho^3\sin\gamma\cos\gamma \: e^{\rho^2(\cos^2\gamma - \sin^2\gamma)}\:d\gamma\:d\rho = \pi^2 \int_0^1 \int_0^{\frac{\pi}{2}} 2\rho^3\sin2\gamma \:e^{\rho^2\cos2\gamma} \:d\gamma\:d\rho$$
since $r,s>0$ by definition. Integrating gives us
$$\pi^2 \int_0^1-\rho \:e^{\rho^2\cos2\gamma}\Bigr|_0^{\frac{\pi}{2}}\:d\rho = \pi^2 \int_0^1 2\rho\:\sinh(\rho^2) \:d\rho = \pi^2(\cosh 1 - 1)$$