Answer
$\frac{1}{3}\cos \frac{1}{x}+C$
Work Step by Step
$\int \frac{\sin(1/x)}{3x^2}dx$
Let $u=\frac{1}{x}=x^{-1}$. Then $du=-x^{-2}dx=-\frac{1}{x^2}dx$, so $\frac{1}{3x^2}dx=-\frac{1}{3}du$
$=\int\sin u*(-\frac{1}{3})du$
$=-\frac{1}{3}\int\sin u du$
$=-\frac{1}{3}(-\cos u+C)$
$=\frac{1}{3}\cos \frac{1}{x}+C$
