Mathematics - Indefinite Integration Question with Solution | TestHub

$\int \frac{\sin x}{\sin (x-\alpha )}\mathrm{dx}=\int \frac{\sin (x-\alpha +\alpha }{\sin (x-\alpha )}\mathrm{dx}$ $=\int \frac{\sin (x-\alpha )\cos \alpha +\cos (x-\alpha )\sin \alpha }{\sin (x-\alpha )}\mathrm{dx}$ $=\int \{\cos \alpha +\sin \alpha \cot (x-\alpha )\}\mathrm{dx}$ $=(\cos \alpha )x+(\sin \alpha )\log \sin (x-\alpha )+\mathrm{C}$ $\therefore \mathrm{A}=\cos \alpha ,\mathrm{B}=\sin \alpha$