Question

The value of \( \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} \, dx \) is:

• A. $2(\sin x + x \cos \alpha) + c$
• B. $2(\sin x - x \cos \alpha) + c$
• C. $2(\cos x + x \sin \alpha) + c$
• D. None of these

Hint:

$\cos \alpha$ is a constant here; do not integrate it as $\sin \alpha$.

Answer 1

The Correct Option is A

Step 1: Concept
Use the identity $\cos 2\theta = 2\cos^{2}\theta - 1$.
Step 2: Analysis
$\frac{(2\cos^{2}x - 1) - (2\cos^{2}\alpha - 1)}{\cos x - \cos \alpha} = \frac{2(\cos^{2}x - \cos^{2}\alpha)}{\cos x - \cos \alpha} = 2(\cos x + \cos \alpha)$. Now integrate: $\int 2(\cos x + \cos \alpha) dx$.
Step 3: Conclusion
$= 2\sin x + 2x \cos \alpha + c = 2(\sin x + x \cos \alpha) + c$.
Final Answer: (A)