$ \begin{align} \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty x^{2k}\,e^{-x^2/2}\,\mathrm{d}x &=\frac2{\sqrt{2\pi}}\int_0^\infty x^{2k-1}\,e^{-x^2/2}\,\mathrm{d}x^2/2\tag{1}\\ &=\frac2{\sqrt{2\pi}}\int_0^\infty (2x)^{k-1/2}\,e^{-x}\,\mathrm{d}x\tag{2}\\ &=\frac{2^{k+1/2}}{\sqrt{2\pi}}\Gamma(k+1/2)\tag{3}\\ &=\frac{(2k)!}{2^kk!}\tag{4} \end{align} $ Explanation:
$(1)$: the integrand is even, so we can double the integral over $(0,\infty)$.
$\phantom{(1)\text{:}}$ Furthermore, $x^{2k-1}\,\color{#C00000}{\mathrm{dx^2/2}} =x^{2k-1}\,\color{#C00000}{x\,\mathrm{d}x}=x^{2n}\,\mathrm{d}x$
$(2)$: Substitute $x\mapsto\sqrt{2x}$
$(3)$: $\Gamma(\alpha)=\int_0^\infty x^{\alpha-1}e^{-x}\,\mathrm{d}x$
$(4)$: $\begin{align}\frac{\Gamma(k+1/2)}{\sqrt\pi} =\frac{\Gamma(k+1/2)}{\Gamma(1/2)} =(k-1/2)(k-3/2)\cdots1/2 =\frac1{2^k}(2k-1)!! =\frac{(2k)!}{4^kk!}\end{align}$