CUET · MATHS · PYQ PAPER 2023
The value of integral \(\int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x\), where \(\alpha\) is a constant is :
- A \(\sin 2 x+\sin 2 \alpha+C\)
- B \(2 x \sin x+\cos \alpha+C\)
- C \(2 \sin x+2 x \sin \alpha+C\)
- D \(2 \sin x+2 x \cos \alpha+C\)
Answer & Solution
Correct Answer
(D) \(2 \sin x+2 x \cos \alpha+C\)
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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