Calculus 8th Edition

$-\frac{sin(\pi/x)}{\pi} +C$
Evaluate the Integral using substitution: $\int \frac{cos(\pi/x)}{x^2}dx$ Substitution Rule: $\int f(g(x))g’(x)dx = \int f(u)du$ $u= \frac{\pi}{x}$ $du =-\frac{\pi}{x^2}$ Since $du$ in the expression is equal to $\frac1{x^2}$ it must be multiplied by $-\frac{1}{\pi}$ Solve the integral in terms of $u$: $\int cos(u)(-\frac1{\pi})du$ $-\frac{1}{\pi}\int cos(u)du$ $-\frac{1}{\pi}sin(u) +C$ $-\frac{sin(u)}{\pi} + C$ Substitute for $u$: $-\frac{sin(\pi/x)}{\pi} +C$