${\int }_{1/3}^3 \frac{1}{x}\sin \left(\frac{1}{x}-x\right)dx$ is equal to

Let $l={\int }_{1/3}^3 \frac{1}{x}\sin \left(\frac{1}{x}-x\right)dx.$ Then,

$\Rightarrow l={\int }_3^{1/3} t\sin \left(t-\frac{1}{t}\right)\times -\frac{1}{t^2}dt,$ where $x=\frac{1}{t}$

$\begin{array}{l}\Rightarrow I={\int }_{1/3}^3 \frac{1}{t}\sin \left(t-\frac{1}{t}\right)dt\\ \Rightarrow I=-{\int }_{1/3}^3 \frac{1}{t}\sin \left(\frac{1}{t}-t\right)dt\\ \Rightarrow I=-1\Rightarrow 2I=0\Rightarrow I=0\end{array}$