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BITSAT 2021
Exam
BITSAT
Year
2021
Subject
Mathematics
Chapter
Indefinite Integration

BITSAT 2021 · Mathematics · Indefinite Integration

If ∫ ( x)/( (x-α)) d x=A x+B (x-α)+C , then value of (A, B) is

If ∫sin⁡xsin⁡(x−α)d\int \frac{\sin x}{\sin (x-\alpha)} \mathrm{d}\mathrm{x}=\mathrm{A}x+Blog⁡sin⁡(x−α)+Cx+\mathrm{B} \log \sin (x-\alpha)+\mathrm{C}, then value of (A,B)(A, B) is

Options

  1. A(−cos⁡α,sin⁡α)(-\cos \alpha, \sin \alpha)
  2. B(cos⁡α,sin⁡α)(\cos \alpha, \sin \alpha)
  3. C(−sin⁡α,cos⁡α)(-\sin \alpha, \cos \alpha)
  4. D(sin⁡α,cos⁡α)(\sin \alpha, \cos \alpha)

Correct answer

B. (cos⁡α,sin⁡α)(\cos \alpha, \sin \alpha)

Step-by-step solution

∫sin⁡xsin⁡(x−α)dx=∫sin⁡(x−α+αsin⁡(x−α)dx=∫sin⁡(x−α)cos⁡α+cos⁡(x−α)sin⁡αsin⁡(x−α)dx=∫{cos⁡α+sin⁡αcot⁡(x−α)}dx=(cos⁡α)x+(sin⁡α)log⁡sin⁡(x−α)+C∴A=cos⁡α,B=sin⁡α\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
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