Aside from some trigonometric substitutions and identities, we will need the following identity, which can be shown using integration by parts twice:
$$
\int_0^{\infty}\cos(\alpha t)e^{-\lambda t}\,\mathrm{d}t=\frac{\lambda}{\alpha^2+\lambda^2}\tag{1}
$$
We will also use the standard arctangent integral:
$$
\int_0^\infty\frac{\mathrm{d}t}{a^2+t^2}=\frac\pi{2a}\tag{2}
$$
Now
$$
\begin{align}
&\left(\int_0^\infty\color{#C00000}{\sin}(x^2) e^{-\lambda x^2}\,\mathrm{d}x\right)^2\\
&=\int_0^\infty\int_0^\infty \color{#C00000}{\sin}(x^2)\color{#C00000}{\sin}(y^2) e^{-\lambda(x^2+y^2)}\,\mathrm{d}y\,\mathrm{d}x\tag{3.1}\\
&=\frac12\int_0^\infty\int_0^\infty \left(\cos(x^2-y^2) \color{#FF0000}{-}\cos(x^2+y^2)\right) e^{-\lambda(x^2+y^2)}\,\mathrm{d}y\,\mathrm{d}x\tag{3.2}\\
&=\frac12\int_0^{\pi/2}\int_0^\infty \left(\cos(r^2\cos(2\phi)) \color{#FF0000}{-}\cos(r^2)\right)e^{-\lambda r^2} \,r\,\mathrm{d}r\,\mathrm{d}\phi\tag{3.3}\\
&=\frac14\int_0^{\pi/2}\int_0^\infty \left(\cos(s\cos(2\phi)) \color{#FF0000}{-}\cos(s)\right) e^{-\lambda s} \,\mathrm{d}s\,\mathrm{d}\phi\tag{3.4}\\
&=\frac14\int_0^{\pi/2}\left(\frac{\lambda}{\cos^2(2\phi)+\lambda^2} \color{#FF0000}{-}\frac{\lambda}{1+\lambda^2}\right)\,\mathrm{d}\phi\tag{3.5}\\
&=\frac12\int_0^{\pi/4} \frac{\lambda}{\cos^2(2\phi)+\lambda^2}\,\mathrm{d}\phi \color{#FF0000}{-}\frac{\lambda\pi/8}{1+\lambda^2}\tag{3.6}\\
&=\frac14\int_0^{\pi/4} \frac{\lambda\,\mathrm{d}\tan(2\phi)} {1+\lambda^2+\lambda^2\tan^2(2\phi)} \color{#FF0000}{-}\frac{\lambda\pi/8}{1+\lambda^2}\tag{3.7}\\
&=\frac14\int_0^\infty\frac{\mathrm{d}t}{1+\lambda^2+t^2} \color{#FF0000}{-}\frac{\lambda\pi/8}{1+\lambda^2}\tag{3.8}\\
&=\frac{\pi/8}{\sqrt{1+\lambda^2}} \color{#FF0000}{-}\frac{\lambda\pi/8}{1+\lambda^2}\tag{3.9}
\end{align}
$$
$(3.1)$ change the square of the integral into a double integral
$(3.2)$ use $2\color{#C00000}{\sin}(A)\color{#C00000}{\sin}(B)=\cos(A-B)\color{#FF0000}{-}\cos(A+B)$
$(3.3)$ convert to polar coordinates
$(3.4)$ substitute $s=r^2$
$(3.5)$ apply $(1)$
$(3.6)$ pull out the constant and apply symmetry to reduce the domain of integration
$(3.7)$ multiply numerator and denominator by $\sec^2(2\phi)$
$(3.8)$ substitute $t=\lambda\tan(2\phi)$
$(3.9)$ apply $(2)$
Finally, take the square root of both sides of $(3)$ and let $\lambda\to0^+$ to get
$$
\int_0^\infty\color{#C00000}{\sin}(x^2)\,\mathrm{d}x=\sqrt{\frac\pi8}\tag{4}
$$
Addendum
I just noticed that the same proof works for
$$
\int_0^\infty\cos(x^2)\,\mathrm{d}x=\sqrt{\frac\pi8}\tag{5}
$$
if each red $\color{#C00000}{\sin}$ is changed to $\cos$ and each red $\color{#FF0000}{-}$ sign is changed to $+$.
$$\int_{0}^{+\infty}\frac{dx}{(x+1)(\pi^2+\log^2 x)}=\int_{0}^{1}\frac{dx}{(x+1)(\pi^2+\log^2 x)}+\int_{0}^{1}\frac{dx}{x(x+1)(\pi^2+\log^2 x)}$$
equals
$$ \int_{0}^{1}\frac{dx}{x(\pi^2+\log^2 x)}\stackrel{x\mapsto e^t}{=}\int_{-\infty}^{0}\frac{dt}{\pi^2+t^2}=\int_{0}^{+\infty}\frac{du}{\pi^2+u^2}=\left[\frac{\arctan(u/\pi)}{\pi}\right]_{0}^{+\infty}=\frac{1}{2}.$$
An overkill is to exploit the integral representation of Gregory coefficients.
Best Answer
Assume $c>0$ for convergence.
$$I = \int_0^{\infty} \dfrac{ dx }{(x^4+c)(x^2+1) } = \frac1{c+1}\int_0^{\infty}\left( \frac{ 1}{x^2+1}- \frac{ x^2-1}{x^4+c }\right) dx \\ = \frac{\pi}{2(c+1)} - \frac{1}{c+1}\int_0^{\infty}\frac{ x^2-1}{x^4+c } dx$$
where
\begin{align} \int_0^{\infty}\frac{ x^2-1}{x^4+c } dx & =\frac{\sqrt c+1}{2\sqrt c} \int_0^{\infty}\frac{ 1-\frac{\sqrt c}{x^2}}{x^2+\frac c{x^2} } dx + \frac{\sqrt c-1}{2\sqrt c} \int_0^{\infty}\frac{ 1+\frac{\sqrt c}{x^2}}{x^2+\frac c{x^2} } dx \\ & =\frac{\sqrt c+1}{2\sqrt c} \int_0^{\infty}\frac{d(x+\frac{\sqrt c}{x})}{(x+\frac {\sqrt c}{x} )^2-2\sqrt c} + \frac{\sqrt c-1}{2\sqrt c} \int_0^{\infty}\frac{d(x-\frac{\sqrt c}{x})}{(x-\frac {\sqrt c}{x} )^2+2\sqrt c} \\ & =0+ \frac{(\sqrt c-1)\pi}{(2\sqrt c)^{3/2}} \end{align}
Thus,
$$I = \frac{\pi}{2(c+1)} - \frac{1}{c+1}\frac{(\sqrt c-1)\pi}{(2\sqrt c)^{3/2}} =\frac{\pi}{2(c+1)} \left( 1- \frac{\sqrt c-1}{\sqrt2 c^{3/4}} \right)$$