If x(x-),dx = Ax+B(x-)+C, then value of (A,B) is:

Question

If $\displaystyle\int \dfrac{\sin x}{\sin(x-\alpha)}\,dx = Ax+B\log\sin(x-\alpha)+C$, then value of $(A,B)$ is:

Accepted Answer

$$\int {{{\sin x} \over {\sin \left( {x - \alpha } \right)}}} dx$$

$$ = \int {{{\sin \left( {x - \alpha  + \alpha } \right)} \over {\sin \left( {x - \alpha } \right)}}} dx$$

$$ = \int {{{\sin \left( {x - \alpha } \right)\cos \alpha  + \cos \left( {x - \alpha } \right)\sin \alpha } \over {\sin \left( {x - \alpha } \right)}}} $$

$$ = \int {\left\{ {\cos \alpha  + \sin \alpha \,\cot \left. {\left( {x - \alpha } \right)} \right\}} \right.} dx$$

$$ = \left( {\cos \alpha } \right)x + \left( {\sin \alpha } \right)\log \,\sin \left( {x - \alpha } \right) + C$$

$$\therefore$$ $$A = \cos \alpha ,$$ $$B = \sin \alpha $$

Solutions · 1