Question

${\int }_{1/2}^2\frac{1}{x}\sin (x-\frac{1}{x})dx=??$

Answer 1

Let $I={\int }_{1/2}^2\frac{1}{x}\sin (x-\frac{1}{x})dx$
Put $x=\frac{1}{t}\Rightarrow dx=-\frac{1}{t^2}dt$
$\therefore I={\int }_2^{1/2}t\sin (\frac{1}{t}-t)(-\frac{1}{t^2})dt={\int }_2^{1/2}\sin (\frac{1}{t}-t)(-\frac{1}{t})dt$
Since $\sin (-\theta )=-\sin \theta$
$I={\int }_{1/2}^2-\frac{1}{t}\sin (t-\frac{1}{t})dt=-I\therefore 2I=0\Rightarrow I=0$