Question
$\int\frac{\cos2\text{x}-\cos2\theta}{\cos\text{x}-\cos\theta}\text{ dx}$ is equal to:
  1. $2(\sin\text{x}+\text{x}\cos\theta)+\text{C}$
  2. $2(\sin\text{x}-\text{x}\cos\theta)+\text{C}$
  3. $2(\sin\text{x}+2\text{x}\cos\theta)+\text{C}$
  4. $2(\sin\text{x}-2\text{x}\cos\theta)+\text{C}$

Answer

  1. $2(\sin\text{x}+\text{x}\cos\theta)+\text{C}$
Solution:
$\text{I}=\int\frac{\cos2\text{x}-\cos2\theta}{\cos\text{x}-\cos\theta}\text{ dx}$
$\text{I}=\int\frac{2\cos^2\text{x}-1-(2\cos^2\theta-1)}{\cos\text{x}-\cos\theta}\text{ dx}$
$\text{I}=\int\frac{2(\cos^2\text{x}-\cos^2\theta)}{\cos\text{x}-\cos\theta}\text{ dx}$
$\text{I}=2\int(\cos\text{x}+\cos\theta)\text{dx}$
$\text{I}=2(\sin\text{x}+\text{x}\cos\theta)+\text{C}$

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