Note that you can convert the integral above to
standard units 5.6.1 so that it is sufficient to show
\begin{equation*}
I = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{ -\frac{z^2}{2} } dz = 1
\end{equation*}
Toward this end, consider \(I^2\) and change the variables to get
\begin{align*}
I^2 & = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{ -\frac{u^2}{2} } du \cdot \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{ -\frac{v^2}{2} } dv\\
& = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{ -\frac{u^2+v^2}{2} } du dv
\end{align*}
Converting to polar coordinates using
\begin{equation*}
du dv = r dr d\theta
\end{equation*}
and
\begin{equation*}
u^2 + v^2 = r^2
\end{equation*}
gives
\begin{align*}
I^2 & = \frac{1}{2 \pi} \int_0^{2 \pi} \int_0^{\infty} e^{ -\frac{r^2}{2} } r dr d\theta\\
& = \frac{1}{2 \pi} \int_0^{2 \pi} -e^{ -\frac{r^2}{2} } \big |_0^{\infty} d\theta\\
& = \frac{1}{2 \pi} \int_0^{2 \pi} 1 \cdot d\theta\\
& = \frac{1}{2 \pi} \theta \big |_0^{2 \pi} = 1
\end{align*}
as desired.