$\int^{\infty}_{-\infty} \int^{\infty}_{-\infty} e^{-\left(2x^2+2xy+2y^2\right)} dx\,dy\,$ 
I need to evaluate $$\displaystyle\int^{\infty}_{-\infty} \int^{\infty}_{-\infty} e^{-\left(2x^2+2xy+2y^2\right)} dx\,dy\,$$

I think I'll need $\displaystyle\int^{\infty}_{-\infty} e^{-x^2} dx\,=\sqrt{\pi}$ . But I'm not able to apply it properly.
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}&\overbrace{\color{#66f}{\large%
\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}
\exp\pars{-\bracks{2x^{2} + 2xy + 2y^{2}}}\,\dd x\,\dd y}}
^{\ds{\dsc{x}=\dsc{\mu + \nu}}\,,\ \dsc{y}=\dsc{\mu - \nu}}
\\[5mm]&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
\exp\pars{-2\bracks{3\mu^{2} + \nu^{2}}}\,\ \overbrace{%
\verts{\partiald{\pars{x,y}}{\pars{\mu,\nu}}}}^{\ds{=\ \dsc{2}}}\,\dd\mu\,\dd\nu
\\[5mm]&=2\int_{-\infty}^{\infty}\exp\pars{-6\mu^{2}}\,\dd\mu
\int_{-\infty}^{\infty}\exp\pars{-2\nu^{2}}\,\dd\nu
\\[5mm]&=2
\bracks{{1 \over \root{6}}\ \overbrace{\int_{-\infty}^{\infty}\exp\pars{-\mu^{2}}\,\dd\mu}^{\dsc{\root{\pi}}}}
\bracks{{1 \over \root{2}}\ \overbrace{\int_{-\infty}^{\infty}\exp\pars{-\nu^{2}}\,\dd\mu}^{\dsc{\root{\pi}}}}
={\pi \over \root{3}}=\color{#66f}{\large{\root{3} \over 3}\,\pi}
\end{align}
A: OK, the problem seems to be completing the square. What one should aim for is to collect the mixed term $xy$ into a squared term. Here I do it with the $x^2$-term:
$$
2x^2+2xy+2y^2=2\bigl(x+\tfrac{1}{2}y\bigr)^2+\frac{3}{2}y^2.
$$
Once that is done, you can change variables (for example $u=\sqrt{2}(x+y/2)$ and $v=\sqrt{3/2}y$. Can you proceed from here?
