real analysis - Compute $int_{-infty}^{infty}int_{-infty}^{infty}e^{-(x^2+(x-y)^2+y^2)}dxdy$

Compute $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+(x-y)^2+y^2)}dxdy$.

I tried to do this by using polar coordinate.
Let $x=r\cos t,\ y=r\sin t$, and then

$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+(x-y)^2+y^2)}dxdy=\int_{0}^{2\pi}\int_{0}^{\infty}e^{-2r^2}e^{2\cos t\sin t}rdrdt=\int_{0}^{2\pi}e^{\sin(2t)}\int_{0}^{\infty}e^{-2r^2}rdrdt=\frac{1}{4}\int_{0}^{2\pi}e^{\sin (2t)}dt.$$
But, I have no idea how to compute $\int_{0}^{2\pi}e^{\sin (2t)}dt$. Please give me some hint or suggestion. Thanks.

Answer

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

It's better to $\underline{\mbox{take advantage}}$ of the $\bbox[5px,#efe]{integrand\ symmetries}$. Namely,

$$\begin{align} &\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp\pars{-\braces{x^{2} + \bracks{x - y}^{2} + y^{2}}}\,\dd x\,\dd y \\[5mm] = &\ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp\pars{-\,{\bracks{x + y}^{2} + 3\bracks{x - y}^{2} \over 2}}\,\dd x\,\dd y \\[5mm] = & \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp\pars{-\,{x^{2} + 3\bracks{x - 2y}^{2} \over 2}}\,\dd x\,\dd y = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp\pars{-\,{x^{2} \over 2} - 6y^{2}}\,\dd x\,\dd y \\[5mm] = &\ \root{2}\,{1 \over \root{6}} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\pars{-x^{2} - y^{2}} \,\dd x\,\dd y = {\root{3} \over 3}\pars{\int_{-\infty}^{\infty}\expo{-x^{2}}\,\dd x}^{2} = \bbx{\ds{{\root{3} \over 3}\,\pi}} \end{align}$$