Question

$\int \frac{\sin \left(x-a\right)}{\sin \left(x-b\right)}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{\sin \left(x-a\right)}{\sin \left(x-b\right)}dx \\ \mathrm{Putting}x-b=t \\ \Rightarrow x=b+t \\ \&dx=dt \\ \therefore I=\int \frac{\sin \left(b+t-a\right)}{\sin t}dt \\ =\int \frac{\sin \left\{\left(b-a\right)+t\right\}}{\sin t}dt \\ =\int \frac{\sin \left(b-a\right)\cos t}{\sin t}+\int \frac{\cos \left(b-a\right)\sin t}{\sin t}dt \\ =\int \sin \left(b-a\right)\cot tdt+\int \cos \left(b-a\right)dt \\ =\sin \left(b-a\right)\ln \left|\sin t\right|+t\cos \left(b-a\right)+C \\ =\sin \left(b-a\right)\ln \left|\sin \left(x-b\right)\right|+\left(x-b\right)\cos \left(b-a\right)+C\left[\because t=x-b\right] \end{gathered}$$