Noting that \begin{aligned}I_{k}-I_{k-2} &=\int \frac{\sin k x-\sin (k-2) x}{\sin x} d x \\&=2 \int \frac{\cos (k-1) x \sin x}{\sin x} d x \\&=2 \int \cos (k-1) x d x \\&=\frac{2}{k-1} \sin (k-1) x+c_{k}\end{aligned} A. For even integer $ n=2m,$ \begin{array}{l} \displaystyle I_{2 k}-I_{2 k-2}=\frac{2}{2 k-1} \sin (2 k-1) x+c_{2 k} \\\displaystyle \sum_{k=1}^{m}\left(I_{2 k}-I_{2 k-2}\right)=2 \sum_{k=1}^{m} \frac{\sin (2 k-1) x}{2 k-1} \\ \displaystyle I_{n}=I_{2 m}=2 \sum_{k=1}^m \frac{\sin (2 k-1) x}{2 k-1}+C_1\end{array} B. For odd integer $ n=2m+1,$ \begin{array}{l}\qquad\qquad \displaystyle \sum_{k=1}^{m}\left(I_{2 k+1}-I_{2 k-1}\right)=2 \sum_{k=1}^{m} \frac{\sin (2 k)}{2 k}\\I_{n}=I_{2 m+1}=I_1+\displaystyle \sum_{k=1}^{m} \frac{\sin (2 k)}{k}+C_{2}=x+\displaystyle \sum_{k=1}^{m} \frac{\sin (2 k)}{k}+C_{2}\end{array}
My Question
How about $$ \int \frac{\sin n x}{\sin m x} d x? $$
