• # question_answer If $\int{\frac{\sin x}{\sin (x-\alpha )}}dx$ $=Ax+B\text{ }\log \text{ }\sin (x-\alpha )+C,$then value of $(A,\text{ }B)$is A) $(sin\text{ }\alpha ,\text{ }cos\text{ }\alpha )$      B)        (b)$(cos\text{ }\alpha ,\text{ }sin\text{ }\alpha \text{)}$ C)        $(-\sin \text{ }\alpha ,\text{ }\cos \text{ }\alpha )$ D)        $(-\cos \text{ }\alpha ,\text{ }\sin \text{ }\alpha )$

Let $l=\int{\frac{\sin x}{\sin (x-\alpha )}}dx$ Put $x-\alpha =t\Rightarrow dx=dt$ $\therefore$  $l=\int_{{}}^{{}}{\frac{\sin (t+\alpha )}{\sin t}dt}$ $l=\int{\frac{\sin t\cos \alpha +\cos t\sin \alpha }{\sin t}}dt$ $l=\int{\cos \alpha dt+}\int{\sin \alpha \frac{\cos t}{\sin t}}dt$ $l=\cos \alpha \int{1dt+}\sin \alpha \int{\frac{\cos t}{\sin t}}dt$ $l=\cos \alpha (t)+\sin \alpha \log \sin t+{{C}_{1}}$ $l=\cos \alpha (x-\alpha )+\sin \alpha \log \sin (x-\alpha )+{{C}_{1}}$ $l=x\cos \alpha +\sin \alpha \log \sin (x-\alpha )-\alpha \cos \alpha +{{C}_{1}}$ $l=x\cos \alpha +\sin \alpha \log \sin (x-\alpha )+C$                                 $(\because \,let\,C=-\alpha \cos \alpha +{{C}_{1}})$ But$\int{\frac{\sin x}{\sin (x-\alpha )}}dx=Ax+B\,\log \,\sin (x-\alpha )+C$ $\therefore$$x\cos \alpha +\sin \alpha \log \sin (x-\alpha )+C$                                 $=Ax+B\,\log \,\sin (x-\alpha )+C$ On comparing, we get $A=\cos \text{ }\alpha ,\text{ }B=\sin \text{ }\alpha$ Alternate Solution $\because$ $\int{\frac{\sin x}{\sin (x-\alpha )}}dx=Ax+B\,\log \,\sin (x-\alpha )+C$ On differentiating both sides w.r.t.$x,$we get $\frac{\sin x}{\sin (x-\alpha )}=A+B\frac{\cos (x-\alpha )}{\sin (x-\alpha )}$ $\Rightarrow$$\sin x=A\sin (x-\alpha )+B\cos (x-\alpha )$ $\Rightarrow$$\sin x=A(\sin x\cos \alpha -\cos x\sin \alpha )$                                 $+B(\cos x\cos \alpha +\sin x\sin \alpha )$ $\Rightarrow$$\sin x=\sin x(A\cos \alpha +B\sin \alpha )$                                 $+\cos x(B\cos \alpha -A\sin \alpha )$ On comparing the coefficients of $\sin x$ and $\cos x$ both sides, we get $A\cos \alpha +B\sin \alpha =1$             ...(i) and        $B\cos \alpha -A\sin \alpha =0$                         ...(ii) On solving Eqs. (i) and (ii), we get $A=cos\alpha ,\text{ }B=sin\alpha$