Find the integration #1

thanks sir Chew-Seong Cheong

$\begin{aligned} I & = \int_0^\infty \frac {\color{#D61F06}\sin^4 x}{x^2} dx & \small \color{#D61F06} \text{See note: } \sin^4 x = \frac {\cos 4x - 4\cos 2x + 3}8 \\ & = \int_0^\infty \frac {\color{#D61F06}\cos 4x - 4\cos 2x + 3}{{\color{#D61F06}8}x^2} dx \\ & = {\color{#3D99F6}\frac 18 \int_0^\infty \frac {\cos 4x}{x^2} dx - \frac 12 \int_0^\infty \frac {\cos 2x}{x^2} dx} + \frac 38 \int_0^\infty \frac 1{x^2} dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = {\color{#3D99F6}- \frac {\cos 4x}{8x}\bigg|_0^\infty - \frac 12 \int_0^\infty \frac {\sin 4x}x dx + \frac {\cos 2x}{2x}\bigg|_0^\infty + \int_0^\infty \frac {\sin 2x}x dx} - \frac 3 {8x} \bigg|_0^\infty & \small \color{#3D99F6} \text{where } \text{Si}(x) = \int_0^x \frac {\sin t}t dt \text{ is the sine integral.} \\ & = -\frac {\color{#D61F06}\cos 4x - 4\cos 2x + 3}{{\color{#D61F06}8}x} \bigg|_0^\infty - \frac 12 {\color{#3D99F6}\text{Si}(\infty)} + {\color{#3D99F6}\text{Si}(\infty)} & \small \color{#D61F06} \text{See note: } \sin^4 x = \frac {\cos 4x - 4\cos 2x + 3}8 \\ & = -\frac {\color{#D61F06}\sin^4 x}x \bigg|_0^\infty + {\color{#3D99F6}\frac \pi 4} & \small \color{#3D99F6} \text{Note that } \text{Si}(\infty) = \frac \pi 2 \\ & = \lim_{x \to 0} \frac {\sin x}x \cdot \sin^3 x - \lim_{x \to \infty} \frac {\sin^4 x}x + \frac \pi 4 \\ & = 0 - 0 + \frac \pi 4 \approx \boxed{0.785} \end{aligned}$

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Note:

$\sin^4 x = \dfrac {(1-\cos 2x)^2}4 = \dfrac {1-2\cos 2x + \cos^2 2x}4 = \dfrac {1-2\cos 2x +\frac 12(1+\cos 4x)}4 = \dfrac {\cos 4x - 4\cos 2x + 3}8$