Can I show $\int^\infty_0\frac{\cos^2x}{x^2}dx=\infty$ by saying $\int^\infty_0\frac{1-\sin^2x}{x^2}dx=\infty-\frac\pi2$? 
$${I=\int^{\infty}_{0}\frac{\cos^{2} x}{x^2}\;dx=\infty}$$

Attempt:
$$\begin{align}&= \int^{\infty}_{0}\frac{1- \sin^{2} x}{x^2}\, dx \tag1 \\[8pt]
&= \infty-\int^{\infty}_{0}\frac{\sin^{2} x}{x^2}\, dx \tag2 \\[8pt]
&= \infty-\frac{\pi}{2} \tag3
\end{align}$$
Hence, $I$ is divergent.

This looks a little better, I'd say.
$$\int_0^\infty\frac{\cos^2x}{x^2}dx>\int_0^{\pi/4}\frac{\cos^2x}{x^2}dx\ge\frac12\int_0^{\pi/4}\frac{dx}{x^2}=\infty$$
