Evaluate the following integrals:
$\int \frac{\sin (\mathrm{x}-\alpha)}{\sin (\mathrm{x}+\alpha)} \mathrm{d} \mathrm{x}$
Add and subtract $\alpha$ in the numerator
$\Rightarrow \int \frac{\sin (x-\alpha+\alpha-\alpha)}{\sin (x+\alpha)} d x$
$\Rightarrow \int \frac{\sin (x+\alpha-2 \alpha)}{\sin (x+\alpha)}$
Numerator is of the form $\sin (A-B)=\sin A \cos B-\cos A \sin B$
Where $A=x+\alpha ; B=2 \alpha$
$\Rightarrow \int \frac{\sin (x+\alpha) \cos (2 \alpha)-\cos (x+\alpha) \sin (2 \alpha)}{\sin (x+\alpha)} d x$
$\Rightarrow \int \frac{\sin (x+\alpha) \cos (2 \alpha)}{\sin (x+\alpha)} d x+\int \frac{\cos (x+\alpha) \sin (2 \alpha)}{\sin (x+\alpha)} d x$
$\Rightarrow \int \cos (2 \alpha) d x+\int \cot (x+\alpha) \sin (2 \alpha) d x$
$\Rightarrow \cos (2 \alpha) \int d x+\sin (2 \alpha) \int \cot (x+\alpha) d x$
As $\int \cot (x) d x=\ln |\sin x|$
$\Rightarrow \cos (2 \alpha) x+\sin (2 \alpha) \ln |\sin (x+\alpha)|$