Answer
-$\frac{sin \frac{2}{x}}{4}$ - $\frac{1}{2x}$ +c
Work Step by Step
u = $\frac{-1}{x}$
du = $\frac{1}{x^{2}}$ dx
$\int$ $\frac{1}{x^{2}}${$cos^{2}$($\frac{1}{x}$)dx = $\int$$cos^{2}$(-u) du
= $\int$$cos^{2}$(u) du
= $\int$ ($\frac{cos 2u}{2}$ + $\frac{1}{2}$ ) du
= $\frac{sin 2u}{4}$ + $\frac{u}{2}$ +c
= -$\frac{sin \frac{2}{x}}{4}$ - $\frac{1}{2x}$ +c
