Integrate the function: $\frac{\sin x}{\sin (x-a)}$

To integrate $I = \int \frac{\sin x}{\sin (x-a)} dx$,we use the substitution method.
Let $t = x-a$,which implies $x = t+a$ and $dx = dt$.
Substituting these into the integral:
$I = \int \frac{\sin (t+a)}{\sin t} dt$
Using the trigonometric identity $\sin (A+B) = \sin A \cos B + \cos A \sin B$:
$I = \int \frac{\sin t \cos a + \cos t \sin a}{\sin t} dt$
Dividing each term by $\sin t$:
$I = \int (\cos a + \cot t \sin a) dt$
Integrating with respect to $t$:
$I = \cos a \int dt + \sin a \int \cot t dt$
$I = t \cos a + \sin a \ln |\sin t| + C_1$
Substituting $t = x-a$ back into the expression:
$I = (x-a) \cos a + \sin a \ln |\sin (x-a)| + C_1$
$I = x \cos a - a \cos a + \sin a \ln |\sin (x-a)| + C_1$
Since $-a \cos a$ is a constant,we can combine it with $C_1$ to form a new constant $C$:
$I = x \cos a + \sin a \ln |\sin (x-a)| + C$