I wonder why under assumption that w>>$\frac{1}{T}$ then $\int_{0}^{T} sin(wt)dt$ is approximately zero?
Since the integral should be like- $\frac{cos(wt)}{w}$ from $0$ to $T$ and after plugging the valued we will end up with :
$\frac{-cos(wT)+1}{w}$
That's a notation commonly employed by physicists to say you need to take limits: the estimate is given by the $\varepsilon-\delta$ definition of the limit, in this case more like $M-\delta$: \begin{equation} \forall M >0\ \exists \delta=\delta(M)>0: wT>M\Rightarrow\frac{-cos(wT)+1}{w}<\delta \end{equation} that is, your approximation gets more and more precise as you get "closer" to $\infty$. In your example of $w=1$ and $T=1000$ you get something of order $1$ as the result of your integral. If you were in a laboratory trying to measure something relating to this formula and wanted something closer to zero, you would just need to set an higher $T$ or $w$ - that's what this formula is telling you. Hope this helps. :-)
(Also, for computing the limit: \begin{equation} \frac{-cos(wT)+1}{w}=\frac{(-cos(wT)+1)}{wT}T \end{equation} even if I guess this wasn't really your problem)