$\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ $\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ 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....

$\bf{My\; Solution::}$ Let $$\displaystyle I = \int_{\frac{1}{2}}^{2}\frac{1}{x}\cdot \sin \left(x-\frac{1}{x}\right)dx$$
Now Let $\displaystyle\left(x-\frac{1}{x}\right) = t\;,$ Then $\displaystyle \left(1+\frac{1}{x^2}\right)dx = dt\Rightarrow \left(x+\frac{1}{x}\right)dx = xdt$ 
and Changing Limits
Now Using $$\displaystyle \left(x+\frac{1}{x}\right)^2 -\left(x-\frac{1}{x}\right)^2=4\Rightarrow \left(x-\frac{1}{x}\right)=\sqrt{\left(x-\frac{1}{x}\right)^2+4}=\sqrt{t^2+4}$$
So Integral $$\displaystyle I = \int_{-\frac{3}{2}}^{\frac{3}{2}}\sin t\cdot \frac{1}{\sqrt{t^2+4}}\cdot \frac{x}{x}dt = \int_{-\frac{3}{2}}^{\frac{3}{2}}\sin t\cdot \frac{1}{\sqrt{t^2+4}}dt$$
So we get $$\displaystyle I = \int_{-\frac{3}{2}}^{\frac{3}{2}}\underbrace{\frac{\sin t}{\sqrt{t^2+4}}}_{\bf{odd\; function}}dt = 0$$
Above we have used the formula $$\displaystyle \int_{-a}^{a}f(x)dx = 0\;,$$ If $f(x)$ is odd function.
