integratin 9

$\begin{aligned} I & = \int_2^4 \frac {\sin^2 x}{x^2} dx \\ & = \int_2^4 \frac {1-\cos (2x)}{2x^2} dx \\ & = \int_2^4 \frac 1{2x^2} dx - \blue{\int_2^4 \frac {\cos (2x)}{2x^2} dx} & \small \blue{\text{Let }u = 2x \implies du = 2\ dx} \\ & = \frac 1{2x} \ \bigg|_4^2 - \blue{\int_4^8 \frac {\cos u}{u^2} du} & \small \blue{\text{By integration by parts}} \\ & = \frac 18 + \frac {\cos u}u \ \bigg|_4^8 + \int_4^8 \frac {\sin u}u du \\ & = \frac 18 + \frac 18 \cos 8 - \frac 14 \cos 4 + \text{ Si }(8) - \text{ Si }(4) & \small \blue{\text{where Si }(\cdot) \text{ denotes the sine integral.}} \\ & \approx \boxed{0.0862} \end{aligned}$

Reference: Sine integral