$\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ $\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ has value equal to
$(A)0\hspace{1cm}(B)\frac{3}{4}\hspace{1cm}(C)\frac{3}{4}\hspace{1cm}(D)2 $
I tried to solve this question by putting $x-\frac{1}{x}=t$ and limits have changed to $\frac{-3}{2}$ to $\frac{3}{2}$ but what to do with remaining $\frac{1}{x}$ and $dx$ in the integration.Please help....

Let $$I(a) = \int^{a}_{\frac{1}{a}}\frac{1}{x}\sin \left(x-\frac{1}{x}\right)dx$$
Where $a>0$
Now $$I'(a) = \frac{1}{a}\sin \left(a-\frac{1}{a}\right)+\sin \left(\frac{1}{a}-a\right)\cdot a \cdot \frac{1}{a^2} = 0$$
So $$I'(a) = 0\Rightarrow I(a) = \mathcal{C}$$
Now put $a=1\;,$ in first equation, We get $I(1) = 0$
So  put  $a = 1$ in $I(a)= \mathcal{C}\;,$ We get $I(1) = 0 = \mathcal{C}$
So we get $\displaystyle I(a) = 0\;,$ Put $a=2\;,$ We get $$I(2) = \int^{2}_{\frac{1}{2}}\frac{1}{x}\sin \left(x-\frac{1}{x}\right)dx = 0$$
