Curve In a Closed Interval with an Infinite Length How can I show that the length of the curve

$$f(x)=\begin{cases}
\sqrt{x}\cos(\pi/x) & \text{ if } x\neq0\\ 
0 & \text{ if } x=0
\end{cases}$$

is infinite on $[0,1]$?
I tried using the arc length formula, but ended up with the very nasty integral:
$$L=\int_0^1\sqrt{1+\left(\frac{x\cos(\pi/x)+2\pi\sin(\pi/x)}{2x^{3/2}}\right)^2}dx.$$
Perhaps the way to go here is to use the definition of arc length I defined in this thread, but I am not certain.
Do you guys have any ideas?
 A: Choosing $x$ so that
$$\pi/x \;\; = \;\; \frac{3 \pi}{2}, \; \frac{4 \pi}{2},\; \frac{5 \pi}{2},\; \frac{6 \pi}{2},\; \frac{7 \pi}{2},\; \frac{8 \pi}{2},\; \frac{9 \pi}{2},\; \frac{10 \pi}{2}, \; \frac{11 \pi}{2},\; \frac{12 \pi}{2},\; \frac{13 \pi}{2},\; ...$$
so that
$$x \;\; =\;\; \frac{2}{3}, \; \frac{2}{4}, \; \frac{2}{5}, \; \frac{2}{6}, \; \frac{2}{7}, \; \frac{2}{8}, \; \frac{2}{9}, \; \frac{2}{10}, \; \frac{2}{11}, \; \frac{2}{12}, \; \frac{2}{13}, \; ... $$
we get
$$\sqrt{x} \cos\left(\pi/x\right) \; \; = \;\; 0, \;\; \sqrt{\frac{1}{2}}, \;\; 0, \;\; -\sqrt{\frac{1}{3}}, \;\; 0, \;\; \sqrt{\frac{1}{4}}, \;\; 0, \;\; -\sqrt{\frac{1}{5}}, \;\; 0, \;\; \sqrt{\frac{1}{6}}, \;\; 0,  \; ...$$
Now, if you plot these points (not necessarily to scale, just make a rough hand sketch), it will be clear that the sum of the lengths of the inscribed segments corresponding to the partition
$$\cal{P}_{n} \;\; =\;\; \left\{1, \;\frac{2}{3}, \; \frac{2}{4}, \; \frac{2}{5}, \; ..., \; \frac{2}{2n}, \; 0\right\}$$
is greater than
$$\sqrt{\frac{1}{2}} \;\; + \;\; \sqrt{\frac{1}{3}} \;\; + \;\; ... \;\; + \;\; \sqrt{\frac{1}{n}}$$
Note that this last sum diverges to $+\infty$ as $n \rightarrow \infty$, as it corresponds to the $p$-series from elementary calculus for $p  =  \frac{1}{2}.$
