Show $\int_\pi^\infty \frac{dx}{x^2 \left( \sin^2(x) \right)^{1/3}}$ is finite using 1st semester measure theory I am studying for a real analysis qualifying exam, and I am completely stuck on this problem.  

Show $$ \int_\pi^\infty \frac{dx}{x^2 \left( \sin^2(x) \right)^{1/3}} $$ is finite.

I tried the following method.  Expand $y^{-\frac{1}{3}}$ about $1$ as $$y^{-\frac{1}{3}} = a_0 + a_1(y-1) + a_2(y-1)^2 + \dots$$ where $a_0 = 1$ and $$a_n = \frac{\left(-1\right)^n}{3^n n!} \prod_{j=1}^n \left( 3j - 2 \right). $$  This converges for $0 < y \leq 2$.  So for $x$ a.e. we may write our function as $$f(x) = \sum_{n=0}^\infty a_n \frac{(\sin^2(x)-1)^n}{x^2} = \sum_{n=0}^\infty \left( -1 \right)^n a_n \frac{\cos^{2n}(x)}{x^2}. $$  
I was hopeful that I could show $f \in L^1([\pi, \infty))$, but the only method I know of is to cross your fingers and hope $$\sum_{n=0}^\infty \left| a_n \right| \int_\pi^\infty \frac{\cos^{2n}(x)}{x^2}dx < \infty. $$  Unfortunately, it can be shown that $\sum_{n=0}^\infty \left| a_n \right| = \infty$, and so my hopes are dashed.  
Question.  Is there a way to salvage my solution?  Or is there another method that uses your standard first semester measure theory techniques?
Aside:  I did see the question here: 
Show that $\int_{\pi}^{\infty} \frac{1}{x^2 (\sin^2 x)^{1/3}} dx$ is finite.
But the solution is not sufficient for my purposes.  
Thank you for your help.    
 A: I think that it will be better find a bound for the integrals
$$\int_{n\pi-c}^{n\pi+c}\frac{dx}{x^2\left(\sin^2(x)\right)^{1/3}}$$
($c$ is some small constant) and proving that numerical series
$$\sum_{n=1}^\infty\int_{n\pi-c}^{n\pi+c}\frac{dx}{x^2 \left( \sin^2(x) \right)^{1/3}}$$
is convergent.
First steps: for $c$ small enough:
$$
\int_{n\pi-c}^{n\pi+c}\frac{dx}{x^2\left(\sin^2(x) \right)^{1/3}}\le
\frac1{(n\pi -c)^2}\int_{n\pi-c}^{n\pi+c}\frac{dx}{\left(\sin^2(x) \right)^{1/3}}\le
\frac2{(n\pi -c)^2}\int_{n\pi-c}^{n\pi+c}\frac{dx}{(x-n\pi)^{2/3}}
$$
($\sin^2(x)\approx(x-n\pi)^2$ when $x\approx n\pi$).
A: I would first show $\int_0^\pi [1/(\sin^2 x)^{1/3}]\, dx < \infty.$ (This follows from the fact that the singularity at $0$ is on the order of $x^{-2/3},$ the singularity at $\pi$ is on the order of $(\pi-x)^{-2/3}.$)
Then observe the integral equals
$$\sum_{n=1}^{\infty}\int_{n\pi}^{(n+1)\pi}\frac{dx}{x^2(\sin^2 x)^{1/3}}\, dx = \sum_{n=1}^{\infty}\int_{0}^{\pi}\frac{dx}{(x+n\pi)^2(\sin^2 x)^{1/3}}\, dx \le \sum_{n=1}^{\infty}\frac{1}{(n\pi)^2}\int_{0}^{\pi}\frac{dx}{(\sin^2 x)^{1/3}}\, dx.$$
