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 $+$.
Best Answer
METHODOLOGY $1$: Using the Laplace Transform
Let $I$ be given by the integral
$$I=\int_0^\infty \frac{\sin^3(x)}{x^2}\,dx$$
Appealing to This Theorem of the Laplace Transform, we first note that for $f(x)=\sin^3(x)$ and $g(x)=\frac1{x^2}$ we have $$\begin{align}\mathscr{L}\{f\}(x)&=\frac{6}{x^4+10x^2+9}\tag 1\\\\ \mathscr{L}^{-1}\{g\}(x)&=x\tag2 \end{align}$$ whence using $(1)$ and $(2)$ in the theorem shows that $$\begin{align} I&=\int_0^\infty \frac{\sin^3(x)}{x^2}\,dx\\\\ &=\int_0^\infty \mathscr{L}\{f\}(x)\mathscr{L}^{-1}\{g\}(x)\,dx\\\\ &=\int_0^\infty \frac{6x}{x^4+10x+9}\,dx\\\\ &=\frac34\int_0^\infty\left(\frac{x}{x^2+1}-\frac{x}{x^2+9}\right)\,dx\\\\ &=\frac38\left.\left(\log(x^2+1)-\log(x^2+9)\right)\right|_{0}^\infty\\\\ &=\frac34\log(3) \end{align}$$ as was to be shown.
METHODOLOGY $2$: Using Feynman's Trick
Let $F(s)$ be given by the integral
$$F(s)=\int_0^\infty \frac{\sin^3(x)}{x^2}e^{-sx}\,dx$$
Differentiating $F(s)$ twice, we find that
$$F''(s)=\frac{6}{s^4+10s^2+9}$$
Integrating $F''(s)$ once reveals
$$F'(s)=\frac34 \arctan(s)-\frac14\arctan(s/3)+C_1$$
Integrating $F'(s)$ we find that
$$F(s)=\frac34 s\arctan(s)-\frac38 \log(s^2+1)-\frac14 s\arctan(s/3)+\frac38\log(s^2+9)+C_1s+C_2$$
Using $\lim_{s\to\infty}F(s)=0$, we find that $C_1=-\pi/4$ and $C_2=0$. Setting $s=0$ yields the coveted result
$$\begin{align} F(0)&=\int_0^\infty \frac{\sin^3(x)}{x^2}\,dx\\\\ &=\frac34\log(3) \end{align}$$
as expected!