$\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi\sigma^2}} e^{(-1/2)((x-\mu)/\sigma)^2} dx = 1$, no polar coordinates

Question

I want to show using the integral below that the area under a normal curve is equal to 1. How can I show that $$\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi\sigma^2}} e^{(-1/2)((x-\mu)/\sigma)^2} dx = 1$$ without polar coordinates, double integrals, or other formulas? Is there a way to only use $$\int_{-\infty}^\infty e^{(-1/2)x^2} dx = \sqrt{2\pi}$$ and $z = (x-\mu)/\sigma$ without any other functions or variables?