Evaluation of an Improper Integral on $\mathbb{R^2}$ 
Problem. Evaluate $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(5x^2-6xy+5y^2)}dxdy$$

My Solution. A hint is given that "Use $\int_{-\infty}^{\infty}e^{-ax^2}dx=\sqrt{\frac{\pi}{a}}$ "
Now,
$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(5x^2-6xy+5y^2)}dxdy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-5(x-\frac{3y}{5})^2}e^{-\frac{16}{5}y^2}dxdy$
But here the range of integration is whole $\mathbb{R^2}$. I know the definition of double integral on a plane bounded region in $\mathbb{R^2}$. So How should I calculate this last integral?
Can I proceed by repeatedly integrate w.r.to $x$ and $y$ like the following:
$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-5(x-\frac{3y}{5})^2}e^{-\frac{16}{5}y^2}dxdy=\int_{-\infty}^{\infty}e^{-\frac{16}{5}y^2}(\int_{-\infty}^{\infty}e^{-5(x-\frac{3y}{5})^2}dx)dy=\int_{-\infty}^{\infty}e^{-\frac{16}{5}y^2}(\sqrt{\frac{\pi}{5}})dy=\sqrt{\frac{\pi}{5}}\sqrt{\frac{\pi}{16/5}}=\frac{\pi}{4}$
Is this right approach? If so how?
 A: $$
5x^2-6xy-5y^2 \equiv 8X^2+2Y^2
$$
This is obtained making a change of variables (rotation) to eliminate the cross product $x y$
now
$$
 \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(5x^2-6xy+5y^2)}dxdy =  \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(8X^2+2Y^2)}dXdY
$$
and then
$$
\int_{-\infty}^{\infty}e^{-8X^2}dX = \frac{\sqrt{\pi}}{2\sqrt 2}\\
\int_{-\infty}^{\infty}e^{-2Y^2}dY = \sqrt{\frac{\pi }{2}}\\
$$
hence
$$
 \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(5x^2-6xy+5y^2)}dxdy =\frac{\pi}{4}
$$
A: Hint: With polar substitution
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(5x^2-6xy+5y^2)}dxdy=\int_{0}^{2\pi}\int_{0}^{\infty}e^{-5r^2+3r^2\sin2\theta}\ r\ drd\theta$$
Can you proceed?
A: $$I=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(5x^2-6xy+5y^2)}dxdy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-5x^2}e^{6xy}e^{-5y^2}dxdy=\left(\int_{-\infty}^{\infty}e^{-5x^2}dx\right)\left(\int_{-\infty}^{\infty}e^{-5y^2}dy\right)\left(\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{6xy}dxdy\right)=\left(\int_{-\infty}^{\infty}e^{-5x^2}dx\right)^2 \left(\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{6xy}dxdy\right)$$
I think this is true, and the first integral left is easy so it would just mean solving the second. Correct me if I'm wrong.
