Answer
$$\int\frac{\cos(\pi/x)}{x^2}dt=-\frac{\sin(\pi/x)}{\pi}+C$$
Work Step by Step
$$A=\int\frac{\cos(\pi/x)}{x^2}dt$$
Let $u=\frac{\pi}{x}$
Then we have $du=\frac{(\pi)'x-\pi(x)'}{x^2}dt=\frac{-\pi}{x^2}dt$. That means $\frac{1}{x^2}dt=-\frac{1}{\pi}du$
Substitute into $A$: $$A=-\frac{1}{\pi}\int\cos udu$$ $$A=-\frac{1}{\pi}\sin u+C$$ $$A=-\frac{\sin(\pi/x)}{\pi}+C$$
