Answer
$\frac{1}{5}\cos \frac{5}{x}+C$
Work Step by Step
$\int \frac{\sin(5/x)}{x^2}dx$
Let $u=\frac{5}{x}=5x^{-1}$. Then $du=-5x^{-2}dx=-\frac{5}{x^2}dx$, so $\frac{1}{x^2}dx=-\frac{1}{5}du$.
$=\int\sin u*(-\frac{1}{5})du$
$=-\frac{1}{5}\int\sin u du$
$=-\frac{1}{5}(-\cos u+C)$
$=\frac{1}{5}\cos \frac{5}{x}+C$