How to solve this integral $$\iint _{\Bbb{R}^2} e^{-(3x+2y)^2 - (4x+y)^2} dx dy$$
I expanded the power terms to get $$I = \iint _{\Bbb{R}^2} e^{-5(5x^2 + y^2 + 4xy)} dx dy = \int e^{-5y^2} \int e^{-25x^2 -20xy} dx dy$$
perhaps this is related to some tricks like of solving $\iint_{\Bbb{R}^2} e^{-x^2} dx$ like of multiplying two integrals and then applying Jacobian?
The obvious change of variables is
$$ u = 3x + 2y, v = 4x+y $$
Then
$$ dudv = \left|\frac{\partial(u,v)}{\partial(x,y)}\right|dxdy = \left|\begin{matrix} 3 & 2 \\ 4 & 1 \end{matrix}\right|dxdy = 5 dxdy $$
And
$$ I = \frac15 \iint_{\Bbb R^2} e^{-(u^2+v^2)}\ dudv $$
Switch to polar coordinates (see solution here) and you're done.