Answer
$$\int\frac{1}{x^2}\cos^2\frac{1}{x}dx=-\frac{1}{2x}-\frac{1}{4}\sin\frac{2}{x}+C$$
Work Step by Step
$$A=\int\frac{1}{x^2}\cos^2\frac{1}{x}dx$$
We set $u=-1/x$.
Then $$du=-\Big(-\frac{1}{x^2}\Big)dx=\frac{1}{x^2}dx$$
Also, $$\frac{1}{x}=-u$$
Therefore, $$A=\int\cos^2(-u)du=\int(\cos(-u))^2du$$
We have, according to trigonometric identities, $\cos(-u)=\cos u$. So $(\cos(-u))^2=\cos^2u$
We also have an identity stating that $\cos^2u=\frac{1+\cos2u}{2}$
Therefore,
$$A=\frac{1}{2}\int(1+\cos2u)du$$ $$A=\frac{1}{2}\Big(u+\frac{1}{2}\sin2u\Big)+C$$ $$A=\frac{1}{2}\Big(-\frac{1}{x}+\frac{1}{2}\sin\Big(-\frac{2}{x}\Big)\Big)+C$$ $$A=\frac{1}{2}\Big(-\frac{1}{x}-\frac{1}{2}\sin\frac{2}{x}\Big)+C$$ (since $\sin(-x)=-\sin x$)$$A=-\frac{1}{2x}-\frac{1}{4}\sin\frac{2}{x}+C$$