∫limits_1 / 2^2 1/x sin ≤ft(x-1/x) d x=

$I=\int\limits_{1 / 2}^2 \frac{1}{x}\sin \left(x-\frac{1}{x}\right) d x$  ...(1)

let $\frac{1}{x}=t$ so $dx=\frac{-1}{t^2}dt$

$⇒I=-\int\limits_{1 / 2}^2\frac{1}{t}\sin \left(t-\frac{1}{t}\right)dt=-\int\limits_{1 / 2}^2 \frac{1}{x}\sin \left(x-\frac{1}{x}\right) d x$   ...(2)

so eq. (1) + eq. (2)

$2I+0⇒I=0$