Calculate the value of:
$${\int }_{1/2}^2\frac{1}{x}\sin \left(x-\frac{1}{x}\right)dx$$

Let $I={\int }_{1/2}^2\frac{1}{x}\sin \left(x-\frac{1}{x}\right)dx$.Use the substitution $x=\frac{1}{t}$, then $dx=-\frac{1}{t^2}dt$.Limits:When $x=1/2$, $t=2$.When $x=2$, $t=1/2$.Substituting into the integral:
$$I={\int }_2^{1/2}t\sin \left(\frac{1}{t}-t\right)\left(-\frac{1}{t^2}\right)dt$$
$$I={\int }_{1/2}^2\frac{1}{t}\sin \left(\frac{1}{t}-t\right)dt$$
Since $\sin (-\theta )=-\sin \theta$:
$$I={\int }_{1/2}^2\frac{1}{t}\left[-\sin \left(t-\frac{1}{t}\right)\right]dt$$
$$I=-I$$
$$2I=0⟹I=0$$