Solution:

It is better to eliminate the denominator, in order to solve these equations.
$\Rightarrow \int \frac{\sin (x-a)}{\sin (x-b)} d x$
Now, add and subtract $b$ in $(x-a)$
$\begin{array}{l}
\Rightarrow \int \frac{\sin (x-a+b-b)}{\sin (x-b)} d x \\
\Rightarrow \int \frac{\sin (x-b+b-a)}{\sin (x-b)}
\end{array}$
The numerator is of the form $\sin (A+B)=\sin A \cos B+\cos A \sin B$
Where $A=x-b ; B=b-a$
$\begin{array}{l}
\Rightarrow \int \frac{\sin (x-b) \cos (b-a)+\cos (x-b) \sin (b-a)}{\sin (x-b)} d x \\
\Rightarrow \int \frac{\sin (x-b) \cos (b-a)}{\sin (x-b)} d x+\int \frac{\cos (x-b) \sin (b-a)}{\sin (x-b)} d x \\
\Rightarrow \int \cos (b-a) d x+\int \cot (x-b) \sin (b-a) d x \\
\Rightarrow \cos (b-a) \int d x+\sin (b-a) \int \cot (x-b) d x \\
\text { As } \int \cot (x) d x=\ln |\sin x| \\
\Rightarrow \operatorname{Cos}(b-a) x+\sin (b-a) \log |\sin (x-b)|
\end{array}$
Hence,
$=\cos (b-a) x+\sin (b-a) \log |\sin (x-b)|+c$, where $c$ is an arbitrary constant.