Convergence of an improper integral with trig function $$\int_0^{ + \infty } {\left( {\frac{x}{{1 + {x^6}{{\sin^2 x}}}}} \right)dx}$$
I'd like your help with see why does??
I tried to compare it to other functions and to change the variables, but it didn't work for me.
Thanks a lot!

We can divide up the ray $[0,\infty)$ into subintervals $[(n-\frac{1}{2})\pi,(n+\frac{1}{2})\pi]$, $n=1,2,\ldots$ (there a little piece $[0,\pi/2]$ left over, which we may ignore).  On each subinterval, we compute
$$
\begin{eqnarray*}
\int_{(n-\frac{1}{2})\pi}^{(n+\frac{1}{2})\pi}\frac{x}{1+x^6\sin^2 x}\,dx&\leq& (n+\frac{1}{2})\pi\int_{(n-\frac{1}{2})\pi}^{(n+\frac{1}{2})\pi}\frac{1}{1+x^6\sin^2 x}\,dx\\
&\leq&(n+\frac{1}{2})\pi\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{1}{1+(n-\frac{1}{2})^6\pi^2\sin^2 u}\,du\\
&\leq&(n+\frac{1}{2})\pi\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{1}{1+4(n-\frac{1}{2})^6u^2}\,du\\
&\leq&(n+\frac{1}{2})\pi\int_{-\infty}^{\infty}\frac{1}{1+4(n-\frac{1}{2})^6u^2}\,du\\
&=&\frac{\pi^2(n+\frac{1}{2})}{2(n-\frac{1}{2})^3}
\end{eqnarray*}
$$
These integrals are decaying like $\frac{1}{n^2}$, so the sum over all $n$ converges.
