Here is my idea: $$\lim_{\varepsilon \to 0^+}\int_{\varepsilon\leq |x|\leq\pi-\varepsilon}\Bigg|\frac{\sin (n+1/2)x}{\sin x/2}\Bigg|dx$$
can be written this \begin{align*} \int_{0}^{\pi}\Bigg|\frac{\sin (n+1/2)x}{\sin x/2}\Bigg|dx&=\int_{0}^{\pi}\Bigg|\frac{\sin nx\cos\frac{x}{2}+\cos nx\sin \frac{x}{2}}{\sin x/2}\Bigg|dx\\ &\leq \int_{0}^{\pi}\Bigg|\frac{\sin nx\cos \frac{x}{2}}{\sin x/2}+\cos nx\Bigg|dx\\ &\leq \int_{0}^{\pi}\Bigg|\frac{\sin nx\cos \frac{x}{2}}{\sin x/2}\Bigg|dx+\pi\\ \end{align*}
This is all what I can do, and I still can not find a constant $C$ independent of $n$ such that $$ \int_{0}^{\pi}\Bigg|\frac{\sin (n+1/2)x}{\sin x/2}\Bigg|dx\leq C. $$ So my method might be wrong.