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

Answer: (A)

$0$

$I={\int }_{\frac{1}{2}}^2\frac{1}{x}\sin \left(x-\frac{1}{x}\right)dx$

$put\frac{1}{x}=t\Rightarrow dx=-\frac{1}{t^2}dt$

$=-{\int }_2^{\frac{1}{2}}t\left(-\frac{1}{t^2}\right)\sin \left(\frac{1}{t}-t\right)dt$

$={\int }_{\frac{1}{2}}^2\frac{1}{t}\sin \left(\frac{1}{t}-t\right)dt$

$=-{\int }_{\frac{1}{2}}^2\frac{1}{t}\sin \left(\frac{1}{t}-t\right)dt$

$I=-{\int }_{\frac{1}{2}}^2\frac{1}{x}\sin \left(x-\frac{1}{x}\right)dx$

$\therefore I=0Hence,theoptionAisthecorrectanswer$