Question

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

• A. $(\sin \alpha ,\cos \alpha )$
• B. $(\cos \alpha ,\sin \alpha )$
• C. $(-\sin \alpha ,\cos \alpha )$
• D. $(-\cos \alpha ,\sin \alpha )$

Answer 1

The correct option is B $(\cos \alpha ,\sin \alpha )$

Let $I=\int \frac{\sin x}{\sin (x-\alpha )}dx$

Put $x-\alpha =t\Rightarrow dx=dt$

$\therefore I=\int \frac{\sin (t+\alpha )}{\sin t}dt=\int \frac{\sin t\cos \alpha +\cos t\sin \alpha }{\sin t}dt$

$=\int \cos \alpha dt+\int \sin \alpha \frac{\cos t}{\sin t}dt$

$=t\cos \alpha +\sin \alpha \log \sin t+c$

$=\cos \alpha (x-\alpha )+\sin \alpha \log \sin (x-\alpha )+c$

$=x\cos \alpha +\alpha \cos \alpha +\sin \alpha \log \sin (x-\alpha )+c$

$=x\cos \alpha +\sin \alpha \log \sin (x-\alpha )+C$

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