$\int \frac{1}{\sin (x-a) \sin (x-b)} d x=$

$\frac{1}{\sin (a-b)} \log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+C$

We have,

$\int \frac{1}{\sin (x-a) \sin (x-b)} d x$

$=\frac{1}{\sin (b-a)} \int \frac{\sin \{(x-a)-(x-b)\}}{\sin (x-a) \sin (x-b)} d x$

$=\frac{1}{\sin (b-a)} \int\{\cot (x-b)-\cot (x-a)\} d x$

$=\frac{1}{\sin (b-a)}\{\log \sin (x-b)-\log \sin (x-a)\}+C$

$=\frac{1}{\sin (b-a)} \log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|+C$

$=\frac{1}{\sin (a-b)} \log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+C$