Prove that $\lim_{N \to \infty} \left( \int_{\pi-\pi/(N+1/2)}^{\pi} \frac{\sin((N+1/2)t)}{\sin(t/2)} dt \right)= 0$

Question

The question is prove that $\lim_{N \to \infty} \int_{\pi-\pi/(N+1/2)}^{\pi} \frac{\sin((N+1/2)t)}{\sin(t/2)} dt = 0$.

I can show that $$\int_{\pi-\pi/(N+1/2)}^{\pi} \frac{\sin((N+1/2)t)}{\sin(t/2)} dt = \int_{\pi-\pi/(N+1/2)}^{\pi} \sin(N t) \frac{\cos(t/2)}{\sin(t/2)} dt + \int_{\pi-\pi/(N+1/2)}^{\pi} \cos(N t) dt$$ and that $ \int_{\pi-\pi/(N+1/2)}^{\pi} \cos(N t) dt$ has limit 0, but working with $\int_{\pi-\pi/(N+1/2)}^{\pi} \sin(N t) \frac{\cos(t/2)}{\sin(t/2)} dt$ is a little harder.

I do know that, for any fixed $M$, $\int_{\pi-\pi/(M+1/2)}^{\pi} \sin(N t) \frac{\cos(t/2)}{\sin(t/2)} dt$ has limit equal to 0 when $N \to \infty$ using some analysis relating to Fourier Coefficients, but proving this result with a moving endpoint is a little more difficult.

Answer 1

$\sin (t/2) \ge \sin (\frac{\pi\left(1 - \frac{2}{N+1}\right)}{2})$ for $t \in \left[\pi - \frac{2\pi}{N+1}, \pi\right]$. Then, $$\left|\int_{\pi - \frac{2\pi}{N+1}}^{\pi} \frac{\sin\left(\frac{N+1}{2} t\right)}{\sin(t/2)} dt \right| \le \int_{\pi - \frac{2\pi}{N+1}}^{\pi} \frac{dt}{\sin \left(\frac{\pi}{2}\left(1 - \frac{2}{N+1}\right)\right)} = \frac{2\pi}{N+1}\frac{1}{\sin \left(\frac{\pi}{2}\left(1 - \frac{2}{N+1}\right)\right)} \rightarrow 0$$