$\text{ Let }\alpha \in \left(0,\frac{\pi }{2}\right)\text{ be fixed. If }\int \frac{\sin (x+\alpha )}{\sin (x-\alpha )}dx=f_1\left(x\right)\cos 2\alpha +f_2\left(x\right)\sin 2\alpha +c\text{ where c constant of integration then }$

$\begin{array}{l}\int \frac{\sin (x+\alpha )}{\sin (x-\alpha )}dx=\int \frac{\sin (x-\alpha +2\alpha )}{\sin (x-\alpha )}dx\text{ Put }x-\alpha =\theta \Rightarrow dx=d\theta \\ =\int \frac{\sin (\theta +2\alpha )}{\sin \theta }d\theta =\cos 2\alpha \int d\theta +\sin 2\alpha \int \frac{\cos \theta }{\sin \theta }d\theta =\cos 2\alpha \cdot (x-\alpha )+\sin 2\alpha \log |\sin (x-\alpha \left)\right|+c\end{array}$