Q65 / 120 · VITEEE 2019 · Mathematics
If $\int \frac{\sin x}{\sin (x-\alpha)} d x=A x+B \log \sin (x-\alpha)+C$, then value of $(A, B)$ is
- A$(-\cos \alpha, \sin \alpha)$
- Correct answerB$(\cos \alpha, \sin \alpha)$
- C$(-\sin \alpha, \cos \alpha)$
- D$(\sin \alpha, \cos \alpha)$
Solution
$$
\begin{array}{l}
\int \frac{\sin x}{\sin (x-\alpha)} d x=\int \frac{\sin (x-\alpha+\alpha)}{\sin (x-\alpha)} d x \\
=\int \frac{\sin (x-\alpha) \cos \alpha+\cos (x-\alpha) \sin \alpha}{\sin (x-\alpha)} d x
\end{array}
$$
$$
\begin{array}{l}
=\int\{\cos \alpha+\sin \alpha \cot (x-\alpha)\} d x \\
=(\cos \alpha) x+(\sin \alpha) \log \sin (x-\alpha)+C \\
\therefore A=\cos \alpha, B=\sin \alpha
\end{array}
$$
Hence, the correct answer is option: 'B'.