Evaluate $\int_\frac12 ^2 \frac1x\tan(x-\frac1x)dx $

Question

$$\int\limits_\frac{1}{2} ^2 \frac{1}{x}\tan\left(x-\frac{1}{x}\right)\mathrm{d}x$$

I have tried substitution and by parts and it seems failed at all. Can anyone give me some hints?

Answer 1

$$I = \int_{\frac{1}{2}}^{2}\frac{1}{x}\tan\left(x - \frac{1}{x}\right)dx = \int_{2}^{\frac{1}{2}}u\cdot \frac{-1}{u^2}\tan\left(\frac{1}{u}-u\right)du = -\int_{\frac{1}{2}}^{2}\frac{1}{u}\tan\left(u - \frac{1}{u}\right)du = -I$$ Where the second step was obtained by letting $\displaystyle x = \frac{1}{u}, dx = \frac{-1}{u^2}du$.

Since $I = -I$, we have $I = 0$.

Answer 2

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 $$I = \int_{-\ln(2)}^{\ln(2)}\underbrace{\tan\left(e^{t}-e^{-t}\right)}_{\bf{Odd\; function}}dt = 0$$

Answer 3

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.