A quick favorite. To evaluate $I = \int_{-\infty}^{\infty} e^{-x^2}\,dx$, square it and switch to polar coordinates:

$$I^2 = \int_{-\infty}^{\infty}\!\!\int_{-\infty}^{\infty} e^{-(x^2+y^2)}\,dx\,dy = \int_0^{2\pi}\!\!\int_0^{\infty} e^{-r^2} r\,dr\,d\theta = \pi.$$

So $I = \sqrt{\pi}$. (Placeholder post — replace with your own.)