CUET · MATHS · PYQ PAPER 2025
\(\int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x\) is equal to
- A \(2 \sin x+2 x \cos \alpha+C\), where \(C\) is an arbitrary constant
- B \(2 \sin x+2 \sin \alpha+C\), where \(C\) is an arbitrary constant
- C \(\frac{\sin 2 x}{2}+x \cos 2 \alpha+C\), where \(C\) is an arbitrary constant
- D \(\frac{\sin 2 x}{2}+\frac{\sin 2 \alpha}{2}+C\), where \(C\) is an arbitrary constant
Answer & Solution
Correct Answer
(A) \(2 \sin x+2 x \cos \alpha+C\), where \(C\) is an arbitrary constant
Step-by-step Solution
Detailed explanation
\( \int \frac{2\cos^2 x - 1 - (2\cos^2 \alpha - 1)}{\cos x-\cos \alpha} d x \) \( \int \frac{2(\cos^2 x - \cos^2 \alpha)}{\cos x-\cos \alpha} d x \) \( \int \frac{2(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x-\cos \alpha} d x \) \( \int 2(\cos x + \cos \alpha) d x \)…
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