Question

If $\int \frac{\sin x}{\sin (x-\alpha )}dx=Ax+B\log |sin(x-\alpha )|+C,$ then the value of $A-B$ at $\alpha =\frac{\pi }{2}$ is

• A. $-1$
• B. $1$
• C. $2$
• D. $0$
• E. $-2$

Answer 1

Answer: (E)

Let $I=\int \frac{\sin x}{\sin (x-\alpha )}dx$ Put $x-\alpha =r\Rightarrow dx=dt$
$\therefore$ $I=\int \frac{\sin (t+\alpha )}{\sin t}dt$
$\Rightarrow$ $I=\int \cos \alpha dt+\int \sin \alpha \frac{\cos t}{\sin t}dt$
$\Rightarrow$ $I=\cos \alpha \int 1dt+\sin \alpha \int \frac{\cos t}{\sin t}dt$
$\Rightarrow$ $I=\cos \alpha (x-\alpha )+\sin \alpha \log \sin (x-\alpha )+C_1$
$\Rightarrow$
$=x\cos \alpha +\sin \alpha \log \sin (x-\alpha )+C$ But $\int \frac{\sin x}{\sin (x-\alpha )}dx$
$=Ax+B\log \sin (x-\alpha )+C$
$\therefore$ $x\cos \alpha +\sin \alpha \log \sin (x-a)+C$
$=Ax+B\log \sin (x-\alpha )+C$ At $\alpha =\frac{\pi }{2},A=0$ and $B=1$
$\therefore$ $A-B=-1$