Question

If $I={\int }_{1/\pi }^{\pi }\frac{1}{x}\cdot \sin (x-\frac{1}{x})dx$ then I is equal to

• A. $0$
• B. $\pi$
• C. $\pi -\frac{1}{\pi }$
• D. $\pi +\frac{1}{\pi }$

Answer 1

The correct option is A $0$

Let $I={\int }_{\frac{1}{\pi }}^{\pi }\frac{1}{x}\cdot \sin (x-\frac{1}{x})dx$
Substitute $x=\frac{1}{t}$
$\therefore I={\int }_{\pi }^{\frac{1}{\pi }}t\sin (\frac{1}{t}-t)(-\frac{1}{t^2})dt$
$=-{\int }_{\frac{1}{\pi }}^{\pi }\frac{1}{t}\sin (t-\frac{1}{t})dt-I\Rightarrow 2I=0\Rightarrow I=0$