How do I determine this integral? $\int_{0}^{+\infty}\sin^2(1/x)\frac{dx}{(4+x^2)^2}$ $$\int_{0}^{\infty}\mathrm dx{\sin^2\left({a\over x}\right)\over (4a^2+x^2)^2}$$
$${\sin^2\left({a\over x}\right)\over (4a^2+x^2)^2}={1-\cos^2\left({a\over x}\right)\over (4a^2+x^2)^2}={1\over 2(4a^2+x^2)^2}-{\cos\left({2a\over x}\right)\over 2(4a^2+x^2)^2}$$
$${1\over 2}\int_{0}^{\infty}\mathrm dx{1\over (4a^2+x^2)^2}-{1\over 2}\int_{0}^{\infty}\mathrm dx{\cos\left({2a\over x}\right)\over (4a^2+x^2)^2}$$
$${\pi\over 8(2a)^3}-{1\over 2}\int_{0}^{\infty}\mathrm dx{\cos\left({2a\over x}\right)\over (4a^2+x^2)^2}$$

$$\int \mathrm dx{1\over (b^2+x^2)^2}={x\over 2b^2(b^2+x^2)}+{1\over 2b^3}\arctan\left({x\over b}\right)+K$$

$$\int_{0}^{\infty}\mathrm dx{\cos\left({2a\over x}\right)\over (4a^2+x^2)^2}$$
Enforcing a substitution of $u=\dfrac{2a}{x}$
$${1\over (2a)^3}\int_{0}^{\infty}\mathrm du {u^2\cos(u)\over (1+u^2)^2}$$
Now this integral is more harder than the original due to the extra $u^2$ at the numerator.
This is an even function, so can be expressed as 
$${1\over 2(2a)^3}\int_{-\infty}^{\infty}\mathrm du {u^2\cos(u)\over (1+u^2)^2}$$
Decomposition of fraction
$${u^2\over (1+u^2)^2}={Au+B\over 1+u^2}+{Cu+B\over (1+u^2)^2}$$
This look like a nightmare, so how do I determine this integral?

The value of $a$ is irrelevant, the $a$-parameter can be removed through a suitable substitution, and by replacing $x$ with $\frac{1}{x}$ the problems boils down to computing
$$ \int_{0}^{+\infty}\frac{\sin^2(x)}{(4x+1/x)^2}\,dx=\frac{1}{16}\int_{0}^{+\infty}\frac{1-\cos(x)}{(x+1/x)^2}\,dx =\frac{1}{16}\left[\frac{\pi}{4}-\int_{0}^{+\infty}\frac{\cos(x)}{(x+1/x)^2}\,dx\right]$$
or 
$$ \text{Re}\int_{-\infty}^{+\infty}\frac{x^2 e^{ix}}{(x^2+1)^2}\,dx\stackrel{\text{Residues}}{=}\color{red}{0} $$
which leads to $\int_{0}^{+\infty}\frac{\sin^2(x)}{(4x+1/x)^2}\,dx = \color{red}{\frac{\pi}{64}}$.
