I'm given the following: $\lim_{n\to \infty}\int_{-\pi}^{\pi}\frac{\sin(n+1/2)t}{\tan(t/2)}dt$.
Upon further simplification I get the following: $\lim_{n\to \infty}\big[\int_{-\pi}^{\pi}\frac{\sin(nt)(\cos(t/2))^2}{\sin(t/2)dt}+\int_{-\pi}^{\pi}\cos(nt)cos(t/2)dt\big]$.
I know by RL Lemma the second integral will go to $0$.
My question is if RL Lemma would also make the first integral go to $0$? Is it possible to conclude that?
$$I_n=2\int_{0}^{\pi} \frac{\sin (n+1/2)}{\tan(x/2)} dx\implies I_n-I_{n-1}=2\int_{0}^{\pi} \frac{\sin(n+1/2)x-\sin(n-1/2)x}{\tan(x/2)}dx$$ $$I_n-I_{n-1}\implies 2\int_{0}^{\pi} 2 \cos nx \cos (x/2) dx= 2\int_{0}^{\pi} [\cos(n+1/2)x+\cos(n-1/2)x] dx=2\left[\frac{\sin(n+1/2)x}{n+1/2}+\frac{\sin(n-1/2)x}{n-1/2}\right]_{0}^{\pi}=2(-1)^n \left[\frac{1}{n+1/2}-\frac{1}{n-1/2} \right]=\frac{2(-1)^{n+1}}{n^2-1/4}.$$ With $I_{0}=8, I_1=8/3$ all $I_n$ can be calcylated recursively as $$I_n=I_{n-1}+\frac{2(-1)^{n+1}}{n^2-1/4}.$$