This is a classic integral from summing Fourier series. I saw the following argument in Katznelson's book I believe. By symmetry we can assume $x > 0$. We divide the integral into $0$ to ${1 \over N}$ and ${1 \over N}$ to $x$ parts. (The second term will be $0$ if $x$ is small enough). Since $|\sin((N + {1 \over 2})t)| \leq (N + {1 \over 2})t$ and $\sin({t \over 2}) \geq C't$ the integrand is bounded by $CN$ and the first term is at most $C$.
For the second term, you can integrate by parts, integrating the $\sin((N + {1 \over 2})t)$ and differentiating the ${1 \over \sin({t \over 2})}$. You get a factor of ${1 \over N + {1 \over 2}}$ from the integration, and the resulting integrand is now bounded by $C{1 \over Nt^2}$. Taking absolute values and integrating from ${1 \over N}$ to $x$ again gives a bound of $C$. The left endpoint term in this integration by parts is bounded by $C{1 \over Nt}$ at $t = {1 \over N}$, so once again you just get a $C$. The right-endpoint term is even smaller. Thus you're done.