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

Solution:

We have, 

$\begin{array}{l}\int \frac{1}{\sin (x-a)\sin (x-b)}dx\\ =\frac{1}{\sin (b-a)}\int \frac{\sin \left\{\right(x-a)-(x-b\left)\right\}}{\sin (x-a)\sin (x-b)}dx\\ =\frac{1}{\sin (b-a)}\int \{\cot (x-b)-\cot (x-a\left)\right\}dx\\ =\frac{1}{\sin (b-a)}\{\log \sin (x-b)-\log \sin (x-a\left)\right\}+C\\ =\frac{1}{\sin (b-a)}\log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|+C\\ =\frac{1}{\sin (a-b)}\log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+C\end{array}$