I wonder why under assumption that \$\omega \gg \frac{1}{T}\$ then \$\int_{0}^{T} \sin(\omega t)dt \approx 0\$?
Since the integral should be like \$\frac{\cos(\omega t)}{w}\$ from \$0\$ to \$T\$ and after plugging the valued we will end up with :
$$\frac{-\cos(\omega T)+1}{\omega}$$
If you are talking about telecommunications, I assume we are talking about high frequencies. If that is the case:
\$−\cos(\omega T)+1\$ ranges from \$0\$ to \$+2\$, if you divide this by a big number you get approximately zero.
To give you an idea: for a frequency around \$1\;\text{kHz}\$ (which is considered "ultra low"), the result will be AT MAXIMUM \$0.002\$.
By increasing the frequency, we're putting more oscillation periods in the integration interval.
Since the integral of a sine over one period is zero, we should only consider the "incomplete" period at the end of the integration interval.
When we increase the frequency, the area of this incomplete period becomes thinner and thinner (explaining the \$\omega\$ in the determinator).
If I plug in some values, I get the following:
\$T = 1\$
\$\omega \rightarrow\$ result
\$10^0 \rightarrow 0.460\$
\$10^1 \rightarrow 0.184\$
\$10^2 \rightarrow 0.001\$
\$10^3 \rightarrow 4.376E-04\$
\$10^4 \rightarrow 1.952E-04\$
\$10^5 \rightarrow 1.999E-05\$
\$10^6 \rightarrow 6.325E-08\$
Now I'm not sure which order of magnitude \$>>\$ signifies and how small the result must be to be considered \$\approx 0\$, but it tends to get zero if it is much larger.
What are the typical values for \$\omega\$ and T you are looking at?
Update (because of the comments):
As FMarazzi has explained quite well there is an upper boundary for the case that \$\cos(\omega T)\$ is -1, so you'll have \$\frac{2}{\omega}\$, which is the absolute maximum you will ever get for any T.
So if you choose the value for T, in a way you get the maximum for a given \$\omega\$ the table turns into:
\$\omega \rightarrow\$ maximum possible value
\$10^0 \rightarrow 2\$
\$10^1 \rightarrow 0.2\$
\$10^2 \rightarrow 0.02\$
\$10^3 \rightarrow 2E-03\$
\$10^4 \rightarrow 2E-04\$
\$10^5 \rightarrow 2E-05\$
\$10^6 \rightarrow 2E-06\$
And so on. I don't know in which context the approximation is used, but as pointed out by the comments it is for communication systems, and my guess would be that those are not about some UART at 9600 baud but something like ethernet or faster things, so \$\omega\$ is in the order of \$10^7\$ or higher, for which the result of the integral gets small and probably doesn't contribute to the other terms of interest.
In the equation as written a larger \$\omega\$ will result on average in a smaller value of the integral but a larger \$T\$ will not.
I suspect more context is needed to properly understand what is meant.
In particular we need to think about what exactly we mean by "\$\approx 0\$". "\$\approx 0\$" should probablly be intepreted as "negligable" but what "negligable" means is highly dependent on context. If there is some related value that increases with increasing values of \$T\$ then it may be that the result of the integral when large \$T\$ is large but \$\omega\$ is small can still be considered negligable.
