Please help to show a function is unbounded and thus not Riemann Integrable I am wondering how I can show the following function is not Riemann integrable.
Since we are on a closed and bounded interval I couldn't use that it is unbounded, etc (think)
$$g(x)=2x\cos\frac{1}{x^2}+\frac{2\sin(\frac{1}{x^2})}{x}$$ if $x \neq 0$, $0$ if $x=0$ considered on [-1,1]
How could I show it is not Riemann integrable?
I was given the suggestion in the comments to consider $$x_{n}=\frac{1}{\sqrt{(2n+1/2)(\pi)}}$$
When I consider the hint and plug in I get
$$2\frac{1}{\sqrt{(2n+1/2)\pi}}\cos\frac{1}{(2n+1/2)\pi}+2\sin\frac{1}{(2n+1/2)\pi} {\sqrt{(2n+1/2)\pi}}$$
However, I still cant figure out how I can use this to show it is not bounded. Can someone please shed light on this part?
But I am not making progress on how this can be used to prove it is not Riemann integrable

In every nbhd of $0$ there is a strictly decreasing positive sequence $(x_n)_{n\in N}$ with $\sin 1/x_{2 n}^2=1$ and  $\sin 1/x_{2 n-1}^2=-1.$ Since  $2 x_n \cos (1/x_n^2)=0$  we have $$g(x_{2 n})=2/x_{2 n}\; \text { and }\; g(x_{2 n-1}=-2/x_{2 n-1}.$$ The variation in $g$ in any nbhd of $0$ is therefore at least $$\sup_{m\in N} \sum_{n=1}^m |f(x_{2 n-1})-f(x_{2 n})|=\sup_{m\in N}\sum_{n=1}^k(2/x_{2 n-1}+2/x_{2 n})\geq \sup_{m\in N}2/x_{2 m}=\infty.$$ Remark:  The term  $2 x \cos 1/x^2$ is bounded in absolute value by $2$ for $x\ne 0$, so if it were replaced by any term bounded (in absolute value) by some $k>0$ we would have $|f(x_{2 m-1})-f(x_{2 m})|\geq 2/ x_{2 m-1}+2/x_{2 m}-2 k >2/x_{2m}-2 k,$ which goes to $\infty$ as $m\to \infty.$ 
