Evaluate $\int_\frac12 ^2 \frac1x\tan(x-\frac1x)dx $ $$\int\limits_\frac{1}{2} ^2 \frac{1}{x}\tan\left(x-\frac{1}{x}\right)\mathrm{d}x$$
I have tried substitution and by parts and it seems failed at all. Can anyone give me some hints?
 A: Let $$I = \int_{1/2}^{2}\frac{1}{x}\tan \left(x-\frac{1}{x}\right)dx$$
Put $x=e^t$, then $t=\ln(x)$ and $dx=e^t\,dt$ hence, changing limits, we get $$I = \int_{-\ln(2)}^{\ln(2)}\underbrace{\tan\left(e^{t}-e^{-t}\right)}_{\bf{Odd\; function}}dt = 0$$
A: Let $$ I = \int_{\frac{1}{2}}^{2}\frac{1}{x}\cdot \tan \left(x-\frac{1}{x}\right)\,dx$$
Now Let $\displaystyle\left(x-\frac{1}{x}\right) = t\;,$ Then $\displaystyle \left(1+\frac{1}{x^2}\right)\,dx = dt\Rightarrow \left(x+\frac{1}{x}\right)\,dx = x\,dt$
and Changing Limits
Now Using $$ \left(x+\frac{1}{x}\right)^2 -\left(x-\frac{1}{x} \right)^2 = 4\Rightarrow \left(x+\frac{1}{x}\right)=\sqrt{\left(x-\frac{1}{x}\right)^2+4}=\sqrt{t^2+4}$$
So Integral $$ I = \int_{-\frac{3}{2}}^{\frac{3}{2}}\tan t\cdot \frac{1}{\sqrt{t^2+4}}\cdot \frac{x}{x} \, dt = \int_{-\frac{3}{2}}^{\frac{3}{2}}\tan t\cdot \frac{1}{\sqrt{t^2+4}} \, dt$$
So we get $$ I = \int_{-\frac{3}{2}}^{\frac{3}{2}} \underbrace{\frac{\tan t}{\sqrt{t^2+4}}}_{\textbf{odd function}} \, dt = 0$$
Above we have used the formula $$\displaystyle \int_{-a}^a f(x) \, dx = 0\;,$$ If $f(x)$ is odd function.
A: $$I = \int_{\frac{1}{2}}^{2}\frac{1}{x}\tan\left(x - \frac{1}{x}\right)dx = \int_{2}^{\frac{1}{2}}u\cdot \frac{-1}{u^2}\tan\left(\frac{1}{u}-u\right)du = -\int_{\frac{1}{2}}^{2}\frac{1}{u}\tan\left(u - \frac{1}{u}\right)du = -I$$
Where the second step was obtained by letting $\displaystyle x = \frac{1}{u}, dx = \frac{-1}{u^2}du$.
Since $I = -I$, we have $I = 0$.
