Question

${\int }_{1/2}^2\frac{1}{x}\sin (x-\frac{1}{x})dx$ has the value equal to

• A. $0$
• B. $\frac{3}{4}$
• C. $\frac{5}{4}$
• D. $2$

Answer 1

The correct option is A $0$

$I={\int }_{\frac{1}{2}}^2\frac{1}{x}\sin (x-\frac{1}{x})dx$
Put $x=\frac{1}{t}\Rightarrow dx=\frac{1}{t^2}dt$
$I={\int }_2^{\frac{1}{2}}t\sin (\frac{1}{t}-t)\left(\frac{-1}{t^2}\right)dt={\int }_2^{\frac{1}{2}}\frac{1}{t}\sin (t-\frac{1}{t})dt$
$=-{\int }_{\frac{1}{2}}^2\frac{1}{t}\sin (t-\frac{1}{t})dt=-I.$
$\Rightarrow 2I=0\Rightarrow I=0$