Prove that integral of $ 1-\cos\left(\frac{1}{x^2}\right) $ is finite I need to prove that 
$$ \int_0^{\infty} \left(1 - \cos\left(\frac{1}{x^2}\right)\right) dx < \infty $$
My attempt: 
$$ \forall x\in[0,\infty] \hspace{1cm} 0 < 1 - \cos\left(\frac{1}{x^2}\right) < 2 \tag{1}. $$ 
Using L'Hôpital's rule I can show that: 
$$ 1 - \cos\left(\frac{1}{x^2}\right) \underset{x \to \infty}{\sim} \frac{1}{2x^4} \tag{2} $$
Which means that $ 1 - \cos\left(\frac{1}{x^2}\right)$ behaves like $ \frac{1}{2x^4} $ when $ x \to \infty $.
So I think that, there exist $N \in \mathbb{N} $ such that: 
$$ \int_0^{\infty} \left(1 - \cos\left(\frac{1}{x^2}\right)\right)dx < \int_0^{N} 2dx + \int_N^{\infty} \frac{1}{x^4}dx < \infty $$
but I'm not sure. I would appreciate any tips or hints.

$\int_0^\infty(1-\cos (1/x^2))dx= \int_0^1(1-\cos (1/x^2))dx+\int_1^\infty(1-\cos (1/x^2))dx\leq 2+\int_1^\infty2\sin^2 \left(\frac{1}{2x^2}\right)dx\leq 2+\int_1^\infty2\sin \left(\frac{1}{2x^2}\right)dx\leq2+\int_1^\infty2 \frac{1}{2x^2}dx \text{ (as } \sin x\leq x \text{ forall } x>0) =2+\int_1^\infty \frac{1}{x^2}dx .$
