Question

\(\int \frac{\cos(2x) - \cos(2\alpha)}{\cos(x) - \cos(\alpha)} \, dx\) is equal to

• A. \(2(\sin(x) - x \cos(\alpha)) + c\)
• B. \(2(\sin(x) - 2x \cos(\alpha)) + c\)
• C. \(2(\sin(x) + x \cos(\alpha)) + c\)
• D. \(2(\sin(x) + 2x \cos(\alpha)) + c\)

Answer 1

The Correct Option is C

\(\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}\)
We can use the identity \(\cos 2\theta = 2 \cos^2 \theta - 1\)
to rewrite cos 2x as: \(cos 2x = 2 cos^2 x - 1 \)

Similarly, \(cos\ 2α\) can be written as: 
\(\cos 2\alpha = 2 \cos^2 \alpha - 1\)
Substituting these expressions into the integrand, we have:
\(\frac{2 \cos^2 x - 1 - 2 \cos^2 \alpha + 1}{\cos x - cos \alpha}\)

Simplifying further: 
\(\frac{2 \cos^2 x - 2 \cos^2 \alpha}{\cos x - cos \alpha}\)
Factoring out a 2 from the numerator: 
\(\frac{2 (\cos^2 x - cos^2 \alpha)}{cos x - cos \alpha}\)
Now, we can factor the numerator further using the difference of squares identity: 
\(\frac{2[(\cos x + \cos \alpha)(\cos x - \cos \alpha)]}{\cos x - \cos \alpha}\)
Canceling out the common factor \((cos x - cos α)\) in the numerator and denominator, we have: 
\(2(\cos x + \cos \alpha)\)
Now we can integrate this expression: 
\(\int 2 (\cos x + \cos \alpha) \, dx\)
Integrating term by term: 
\(2 \int \cos x \, dx + 2 \int \cos \alpha \, dx\)
The integral of cos x with respect to x is sin x, and the integral of a constant (cos α) with respect to x is the constant times x: 
\(2 \sin x + 2 (\cos \alpha) x + c\) where c is the constant of integration. 

Hence, the integral \(\int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} \, dx\) is equal to \(2 \sin x + 2 (\cos \alpha) x + c\), which corresponds to option (C) \(2(\sin x + x \cos \alpha) + c\)