$\int \frac{\sin x+\cos x}{\sin (x-\alpha)} d x$ is equal to

$(\cos \alpha-\sin \alpha)(x-\alpha)+(\cos \alpha+\sin \alpha) \log |\sin (x-\alpha)|+C$

We have,

$\int \frac{\sin x+\cos x}{\sin (x-\alpha)} d x$

$=\int \frac{\sin (t+\alpha)+\cos (t+\alpha)}{\sin t} d t$, where $t=x-\alpha$

$=\int(\cos \alpha+\sin \alpha \cot t+\cos \alpha \cot t-\sin \alpha) d t$

$=(\cos \alpha-\sin \alpha) t+(\sin \alpha+\cos \alpha) \log |\sin t|+C$

$=(\cos \alpha-\sin \alpha)(x-\alpha)+(\cos \alpha+\sin \alpha) \log |\sin (x-\alpha)|+C$