Calculate $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} dx dy \exp(-5x^2+8xy-5y^2)$ I need to evaluate the following integral
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} dx dy \exp(-5x^2+8xy-5y^2)$$
I know the normal double integrals but this one seem different, moreover, I'm familiar with $$\int_{-\infty}^{\infty} \exp(-x^2)dx = \sqrt\pi$$ but I cannot figure out the polar substitution for $-5x^2+8xy-5y^2$ expression. I tried substituting $x^2+y^2$ at $-5(x^2+y^2)$ with $r^2$ and $8xy$ with $r^2\sin\theta\cos\theta$. And then I got even more lost.
Can someone please help me figure this question out and is there any general solution for such problems? And also can someone provide me a link to similar problems and their solutions in order for me to understand better. Thank you!

To begin, we complete the square to notice that
$$-5x^{2}+8xy-5y^{2}=-5\left(x-\frac{4}{5}y\right)^{2}-\frac{9}{5}y^{2}$$
so we get that
\begin{align*}
&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp(-5x^{2}+8xy-5y^{2})dxdy\\
&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\left(-5\left(x-\frac{4}{5}y\right)^{2}-\frac{9}{5}y^{2}\right)dxdy.\tag{1}
\end{align*}
For any value of $y$ it holds that
$$\int_{-\infty}^{\infty}\exp\left(-5\left(x-\frac{4}{5}y\right)^{2}\right)dx=\int_{-\infty}^{\infty}\exp\left(-5x^{2}\right)dx$$
because of the substiuttion $u=x-\frac{4}{5}y$, so substituting back into (1) we get that
\begin{align*}
&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp(-5x^{2}+8xy-5y^{2})dxdy\\
&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\left(-5x^{2}-\frac{9}{5}y^{2}\right)dxdy\\
&=\left(\int_{-\infty}^{\infty}\exp\left(-5x^{2}\right)dx\right)\left(\int_{-\infty}^{\infty}\exp\left(-\frac{9}{5}y^{2}\right)dy\right)
\end{align*}
Now making the subs $u=\sqrt{5}x$ and $u=\frac{3}{\sqrt{5}}y$ we get that
\begin{align*}
&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp(-5x^{2}+8xy-5y^{2})dxdy\\
&=\left(\frac{1}{\sqrt{5}}\int_{-\infty}^{\infty}\exp\left(-x^{2}\right)dx\right)\left(\frac{\sqrt{5}}{3}\int_{-\infty}^{\infty}\exp\left(-y^{2}\right)dy\right)\\
&=\left(\frac{\sqrt{\pi}}{\sqrt{5}}\right)\left(\frac{\sqrt{5\pi}}{3}\right)\\
&=\frac{\pi}{3}
\end{align*}
So we are done
