Should we really integrate it? 2

$\begin{aligned} I & = \int_{-\infty}^\infty \frac{\sin^2 4x}{(2x)^2}\, dx \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{As the integrand is even}} \\ & = \int_0^\infty \frac{\sin^2 4x}{2x^2}\, dx \\ & = \frac{1}{4} \int_0^\infty \frac{1-\cos 8x}{x^2}\, dx \quad \quad \quad \quad \quad \quad \quad \quad \, \, \small \color{#3D99F6}{\text{By integration by parts}} \\ & = \frac{1}{4} \left[ \frac {\cos 8x -1}{x} \right]_0^\infty + 2 \int_0^\infty \frac{\sin 8x}{x} \, dx \quad \quad \small \color{#3D99F6}{\text{Let }u = 8x \implies du = 8 \, dx} \\ & = 0 + 2 \color{#3D99F6} {\int_0^\infty \frac{\sin u}{u} \, du} \quad \quad \quad \quad \quad \quad \quad \quad \quad \, \, \small \color{#3D99F6}{\text{Dirichlet integral}} \\ & = 2 \left( \color{#3D99F6}{\frac{\pi}{2}} \right) \\ & = \boxed{\pi} \end{aligned}$

Did it using Duality property of Fourier Transform.