Question

If $\int \frac{\sin x}{\sin (x-\alpha )}dx=Ax+B\log |\sin (x-\alpha )|+C$, then

• A. $A=\sin \alpha$
• B. $B=\cos \alpha$
• C. $A=\cos \alpha$
• D. $B=\sin \alpha$

Answer 1

Answer: (C), (D)

The correct options are
C $A=\cos \alpha$
D $B=\sin \alpha$

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

$=\sin \left(\alpha \right)\ln \left|\sin \right(x-\alpha \left)\right|+\cos \left(\alpha \right).x+C$

Hence, options 'C' and 'D' are correct.