Prove $\displaystyle f(x)=\frac{sin x}{1+x^2}$ is uniformly continuous over R

Stuck on this one

So I'm trying to show for appropiate $\displaystyle \delta$, and $\displaystyle |x-a|<\delta$, that $\displaystyle |f(x)-f(a)|<\epsilon$

So

$\displaystyle |\frac{sin x}{1+x^2}-\frac{sin a}{1+a^2}|

$

$\displaystyle =|\frac{sinx-sina+a^2sinx-x^2sina}{(1+x^2)(1+a^2)}|

$

Known that $\displaystyle |x-a|\geq{|sinx-sina|}$

Using triangle inequality and considering the numerator only

$\displaystyle |sinx-sina+a^2sinx-x^2sina|\leq{|x-a|+ |a^2sinx-x^2sina|}

$

And here I'm stuck.

Hints please and thank you