$$I_{k,m}(n)=\int_{0}^{1}u^k\cot{\frac{\pi(1-u)}{m}}\sin{\frac{2\pi n(1-u)}{m}}\,du$$

Conjecture: $\lim_{n\rightarrow\infty} I_{k,m}(n)=m/2$, for $m\geq 1$ and any $k\in\{1,2,3,\ldots\}$.

$$I_{k,m}(n)=\int_{0}^{1}u^k\cot\left({\frac{\pi(1-u)}{m}}\right)\sin\left({\frac{2\pi n(1-u)}{m}}\right)\,du$$

$$\lim_{n\rightarrow\infty} I_{k,m}(n)=\frac{1}{2}m, \;\;\text{for}\;\; m\geq 1\;\; \text{and any} \;\;k\in\{1,2,3,\ldots\}.$$

No proof (yet), but the plot below gives the numerical evidence. It is a collection of graphs of $I_{k,m}(n)$ as a function of $m$, for fixed $n=500$ and $k$ taking the values 1,2,3,4,5.

$$I_{k,m}(n)=\int_{0}^{1}u^k\cot{\frac{\pi(1-u)}{m}}\sin{\frac{2\pi n(1-u)}{m}}\,du$$

Conjecture: $\lim_{n\rightarrow\infty} I_{k,m}(n)=m/2$, for $m\geq 1$ and any $k\in\{1,2,3,\ldots\}$.

$$I_{k,m}(n)=\int_{0}^{1}u^k\cot{\frac{\pi(1-u)}{m}}\sin{\frac{2\pi n(1-u)}{m}}\,du$$

Conjecture: $\lim_{n\rightarrow\infty} I_{k,m}(n)=m/2$, for $m\geq 1$ and any $k\in\{1,2,3,\ldots\}$.