Here are certain weighted Gaussian integrals I have encountered for which numerical computation reassures equality.
Question. Is this true? If so, is there an underlying transformation or just a proof? $$\int_0^{\infty}x^2e^{-x^2}\frac{{dx}}{\cosh\sqrt{\pi}x} =\frac14\int_0^{\infty}e^{-x^2}\frac{dx}{\cosh\sqrt{\pi}x}.$$
UPDATE
The integral in T. Amdeberhan's question was taken from his own paper that was written 10 years earlier before he posted his question: https://arxiv.org/abs/0808.2692 A dozen integrals: Russell-style. Thus it seems that T. Amdeberhan knew the answer to the question he asked. Why did he ask it then?
I provide a screenshot below for the reader's convenience (integral number $7$):
At the end of the paper, the authors provide a sketch of proof:
OLD ANSWER
Note that the following functions are self-reciprocal $\sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\cos ax dx=f(a)$ (see this MO post for a list of such functions): $$ \frac{1}{\cosh\sqrt{\frac{\pi}{2}}x},\quad e^{-x^2/2}.\tag{1} $$ It was proved by Hardy (Quarterly Journal Of Pure And Applied Mathematics, Volume 35, Page 203, 1903) and Ramanujan (Ramanujan's Lost Notebook, part IV, chapter 18) that for two self-reciprocal functions $f$ and $g$ we have $$ \int_0^\infty f(x)g(\alpha x) dx=\frac{1}{\alpha}\int_0^\infty f(x)g(x/\alpha) dx. $$ The formal agument is as follows but it can be made rigorous for certain type of functions $$ \int_0^\infty f(x)g(\alpha x) dx=\int_0^\infty f(x)dx\cdot \sqrt{\frac{2}{\pi}}\int_0^\infty g(y) \cos(\alpha x y)dy=\\ \int_0^\infty g(y)dy \cdot \sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\cos(\alpha x y)dx=\int_0^\infty g(y)f(\alpha y) dy=\\ \frac{1}{\alpha}\int_0^\infty f(x)g(x/\alpha) dx. $$ For functions in $(1)$ which decay rapidly at $x\to\pm\infty$ it is certainly true. So we obtain an identity due to Hardy and Ramanujan $$ \int_{0}^{\infty} \frac{e^{-x^{2}/2}}{\cosh{\sqrt{\frac{\pi}{2}}\alpha{x}}} {dx} = \frac{1}{\alpha} \int_{0}^{\infty} \frac{e^{-x^{2}/2}}{\cosh{\sqrt{\frac{\pi}{2}}\frac{x}{\alpha}}}{dx}. $$ After some simplifications it becomes $$ \int_{0}^{\infty} \frac{e^{-\alpha^2x^{2}}}{\cosh{\sqrt{{\pi}}{x}}} {dx} = \frac{1}{\alpha} \int_{0}^{\infty} \frac{e^{-x^{2}/\alpha^2}}{\cosh{\sqrt{{\pi}}{x}}}{dx}. $$ To complete the proof differentiate this with respect to $\alpha$ and then set $\alpha=1$.
This is a generous explanation of Lucia's comment above.
The functions $$\mathcal H_n(x)=\frac{2^{1/4}}{(2^n n!)^{1/2}}H_n(\sqrt{2\pi}\; x) e^{-\pi x^2}$$ form an orthonormal system in $L^2(\textbf{R})$. Here $H_n(x)$ are the usual Hermite polynomials defined by $$e^{2xz-z^2}=\sum_{n=0}^\infty \frac{H_n(x)}{n!}z^n,\qquad |z|<\infty.$$ The functions are eigenfunctions for the usual Fourier transform so that $$\int_{-\infty}^{+\infty}\mathcal H_n(t)e^{-2\pi i x t}\,dt=(-i)^n \mathcal H_n(x).$$
It follows that $L^2(\textbf{R})$ is a direct sum of four subspaces. In each of these subspaces the Fourier transform is just multiplication by $1$, $i$, $-1$, $-1$ respectively.
It is well known that the function $1/\cosh\pi x$ is invariant by the Fourier transform $$\int_{-\infty}^{+\infty}\frac{e^{-2\pi i x\xi}}{\cosh\pi x}\,dx= \frac{1}{\cosh\pi \xi}.$$ Therefore this function is in the span of the functions $\mathcal H_{4n}(x)$, and therefore is orthogonal to any other eigenfunction. In particular $$\int_{-\infty}^{+\infty}\frac{\mathcal H_2(x)}{\cosh\pi x}\,dx=0.$$ Since $H_2(x)=-2+4x^2$ it follows that $$\int_{-\infty}^{+\infty}\frac{(8\pi x^2-2)e^{-\pi x^2}}{\cosh\pi x}\,dx=0.$$ Putting $x/\sqrt{\pi}$ instead of x $$\int_{-\infty}^{+\infty}\frac{(8 x^2-2)e^{-x^2}}{\cosh\sqrt{\pi} x}\,\frac{dx}{\sqrt{\pi}}=0.$$ This is equivalent to your equation.
