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!
Best Answer
$\sin(x) = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \ldots$ So look at $\displaystyle\int_0^\infty e^{-x^2}x^{2k+1}dx = \frac12\int_0^\infty e^{-t}t^kdt = \dfrac{k!}2$ and you are done. Note: this is more elementary than the proposed method. Let me try the other way.