I need ideas for solve this improper integral, i know is hard and is a bonus for my analysis course, so i would really appreciate your help, thanks
$\int_{1}^{\infty}\dfrac{x\sin(2x)}{x^2+3}dx$
Hint: $\begin{cases} \sin(\theta)\geq \dfrac{2\theta}{\pi},& 0\leq\theta\leq \dfrac{\pi}{2}\\\\\sin(\theta)\geq \dfrac{-2\theta}{\pi}+2,& \dfrac{\pi}{2}\leq\theta\leq{\pi}\end{cases}$
In order to bound the integral
$\int_{0}^{\pi} e^{-2R\sin(\theta)}d\theta$
I don't know a nice and beauty approach in order to attack this properly to obtain a closed answer, so....