We need to calculate
$$
J=\int _0^\infty e^{ix^2}\,dx
$$
because $\int_0^\infty \cos^2(x)\,dx$ is it's real part, and $\int_0^\infty \sin^2(x)\,dx$.
So, consider $\int_\Gamma e^{iz^2}\,dz=0$ due to Cauchy theorem. This integral consist of three terms: integral along horizontal segment $=\int_0^R e^{ix^2}\,dx$, integral over sloped segment $=-\int_0^R e^{-t^2} \frac{(1+i)}{\sqrt 2}dt$ as we plugged $z=\frac{(1+i)}{\sqrt 2} t$. Finally, consider integral over arc:
$$
\int_0^{\pi /4} e^{iR^2e^{2it}} Re^{it}\,dt=\int_0^{\pi/4} e^{-R^2 \sin(t)+R^2 i \cos (t)+it} R\,dt.$$
Its absolute value does not exceed
$$
\int_0^{\pi/4} e^{-R^2 \sin(t)} R\,dt\le 2\int_0^{\pi/4} e^{-R^2 \sin(t)}\cos(t) R\,dt\le 2R^{-1} e^{-R^2\sin(t)}|^{t=0}_{t=\pi/4}\le 2R^{-1}
$$
which tends to $0$ for $R\to \infty$.
Then $J=\int_0^\infty e^{-t^2} \frac{(1+i)}{\sqrt 2}dt$ and we use Q20
