Q44 / 125 · BITSAT 2021 · Mathematics

If $\int \frac{\sin x}{\sin (x-\alpha)} \mathrm{d} \mathrm{x}=\mathrm{A} x+\mathrm{B} \log \sin (x-\alpha)+\mathrm{C}$, then value of $(A, B)$ is

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

Solution

$\int \frac{\sin x}{\sin (x-\alpha)} \mathrm{dx}=\int \frac{\sin (x-\alpha+\alpha}{\sin (x-\alpha)} \mathrm{dx}$ $=\int \frac{\sin (x-\alpha) \cos \alpha+\cos (x-\alpha) \sin \alpha}{\sin (x-\alpha)} \mathrm{d} \mathrm{x}$ $=\int\{\cos \alpha+\sin \alpha \cot (x-\alpha)\} \mathrm{d} \mathrm{x}$ $=(\cos \alpha) x+(\sin \alpha) \log \sin (x-\alpha)+\mathrm{C}$ $\therefore \mathrm{A}=\cos \alpha, \mathrm{B}=\sin \alpha$

Hence, the correct answer is option: 'B'.

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