If $f(x) = { x^2 \sin \frac 1 x \text{ for } 0 < x \leq 1}$ and $f(0)=0$, prove $f$ is rectifiable. If $f(x) = { x^2 \sin \frac 1 x \text{ for } 0 < x \leq 1}$ and $f(0)=0$, prove $f$ is rectifiable.
I tried calculating the length, but couldn't do it.  The actual integral should be $\int _0^1\sqrt{1+\left(2x\sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)\right)^2}\:dx$ but its too hard.  How to prove $f$ is rectifiable?  Also anyone have any ideas how to do that integral?

You don't have to calculate it, you just have to show it's finite. Hint: 
$$(2x\sin(1/x)-\cos(1/x))^2\le 4$$
