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MHT CET · Maths · Indefinite Integration

\(\int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} \mathrm{d} x=\)

  1. A \(2 \cos x+2 x \cos \alpha+c \quad\), where \(c\) is the constant of integration.
  2. B \(2 \cos x-2 x \cos \alpha+c \quad\), where \(c\) is the constant of integration.
  3. C \(2 \sin x+2 x \cos \alpha+c, \quad\) where \(c\) is the constant of integration.
  4. D \(2 \sin x+2 x \sin \alpha+c \quad\), where \(c\) is the constant of integration.
Verified Solution

Answer & Solution

Correct Answer

(C) \(2 \sin x+2 x \cos \alpha+c, \quad\) where \(c\) is the constant of integration.

Step-by-step Solution

Detailed explanation

\(\int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} \mathrm{d} x = \int \frac{(2\cos^2 x - 1)-(2\cos^2 \alpha - 1)}{\cos x-\cos \alpha} \mathrm{d} x \) \(= \int \frac{2(\cos^2 x - \cos^2 \alpha)}{\cos x-\cos \alpha} \mathrm{d} x \)