Substituting $x\mapsto x\sqrt{2k/\pi}$
$$
\sqrt{\frac2\pi}\sum_{k=1}^\infty\frac1{k^2}\int_0^1\cos\left(kx^2\right)\,\mathrm{d}x\tag{1}
$$
Let's compute
$$
\begin{align}
\sum_{k=1}^\infty\frac{\sin(kx)}{k}
&=\mathrm{Im}\left(\sum_{k=1}^\infty\frac{e^{ikx}}{k}\right)\\
&=\mathrm{Im}\left(-\log\left(1-e^{ix}\right)\right)\\
&=-\arg\left(1-e^{ix}\right)\\
&=\arctan\left(\frac{\sin(x)}{1-\cos(x)}\right)\\
&=\arctan\left(\frac{2\sin(x/2)\cos(x/2)}{2\sin^2(x/2)}\right)\\
&=\arctan(\cot(x/2))\\
&=\frac{\pi-x}2\qquad\text{for }x\in(0,2\pi)\tag{2}