Integration in polar coordinates?

Question

Given $$ A=\begin{pmatrix} a & b \\b & c \end{pmatrix}, x=(x_1,x_2), (Ax,x)>0 $$ and $$(x,y)=x_1\cdot y_1+x_2\cdot y_2$$ I'm trying to prove that $$ \int_{-\infty}^\infty \int_{-\infty}^\infty \! e^{-(Ax,x)} \, \mathrm{d}x_1 \mathrm{d}x_2 = \frac \pi{\sqrt{\det(A)}} $$ as the n=2 case of $$ \int_{-\infty}^\infty \int_{-\infty}^\infty...\int_{-\infty}^\infty \! e^{-(Ax,x)} \, \mathrm{d}x_1 \mathrm{d}x_2...\mathrm{d}x_n = \frac {\pi^{n/2}}{\sqrt{\det(A)}} $$ (where A is a nxn symmetric matrix). What worked in the case n=1, $$ I=\int_{-\infty}^\infty \! e^{-x^2} \, \mathrm{d}x = \sqrt{\pi} $$ was taking $$I^2$$ and substituning polar coordinates. As kindly hinted in the comments, we can diagonalize the quadratic form. There is an orthogonial matrix $$ P:P^{-1}AP=\begin{pmatrix} l_1 & 0 \\0 & l_2\end{pmatrix}$$. Then: $$ I=\int_{-\infty}^\infty \int_{-\infty}^\infty \! e^{-(l_1z_1^2+l_2z_2^2)} \, \mathrm{d}z_1 \mathrm{d}z_2 =\int_0^{2π} \int_{0}^\infty \! e^{-(l_1r^2cos^2θ+l_2r^2sin^2θ)} \, \mathrm{d}r \mathrm{d}θ = \int_{0}^\infty e^{-r^2}\mathrm{d}r \int_{0}^{2π} \! e^{l_1cos^2θ+l_2sin^2θ} \, \mathrm{d}θ = \frac{1}{2} \int_{0}^{2π} \! e^{l_1cos^2θ+l_2sin^2θ} \, \mathrm{d}θ $$ This is as far as I can go. Any help would be appreciated.