Practicing Success
If $\int \frac{\sin x}{\sin (x-\alpha)} d x =A x+B \log \sin (x-\alpha)+C$, then the value of $(A, B)$, is |
$(-\cos \alpha, \sin \alpha)$ $(\cos \alpha, \sin \alpha)$ $(-\sin \alpha, \cos \alpha)$ $(\sin \alpha, \cos \alpha)$ |
$(\cos \alpha, \sin \alpha)$ |
We have, $\int \frac{\sin x}{\sin (x-\alpha)} d x=\int \frac{\sin (x-\alpha+\alpha)}{\sin (x-\alpha)} d x$ $=\int \frac{\sin (x-\alpha) \cos \alpha+\cos (x-\alpha) \sin \alpha}{\sin (x-\alpha)} d x$ $=\int \cos \alpha d x+\sin \alpha \int \cot (x-\alpha) d x$ $=x \cos \alpha+\sin \alpha \log |\sin (x-\alpha)|+C$ ∴ $A=\cos \alpha, B=\sin \alpha$ $\Rightarrow (A, B)=(\cos \alpha, \sin \alpha)$ |