Answer
$-\frac{sin(\pi/x)}{\pi} +C$
Work Step by Step
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 $u$:
$-\frac{sin(\pi/x)}{\pi} +C$
