Question

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

• A. \(2 \cos x + 2x \cos \alpha + c\), where c is the constant of integration.
• B. \(2 \cos x - 2x \cos \alpha + c\), where c is the constant of integration.
• C. \(2 \sin x + 2x \cos \alpha + c\), where c is the constant of integration.
• D. \(2 \sin x + 2x \sin \alpha + c\), where c is the constant of integration.

Hint:

When an integrand contains \(\cos 2x - \cos 2\alpha\), try factorizing it first. It often cancels the denominator immediately.

Answer 1

The Correct Option is C

Concept:
Use the identity: \[ \cos 2x - \cos 2\alpha = 2(\cos x - \cos \alpha)(\cos x + \cos \alpha) \] This lets the denominator cancel directly. ip

Step 1: Simplify the integrand.
\[ \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} = \frac{2(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} \] \[ =2(\cos x + \cos \alpha) \] ip

Step 2: Integrate term by term.
\[ \int 2(\cos x + \cos \alpha)\,dx = 2\int \cos x\,dx + 2\int \cos \alpha\,dx \] \[ =2\sin x + 2x\cos \alpha + c \] ip Hence, the correct answer is:
\[ \boxed{(C)\ 2 \sin x + 2x \cos \alpha + c} \]