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

Answer: (C)

$2(\sin(x) + x \cos(\alpha)) + c$

Answer 2

$\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}$
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) $2(\sin x + x \cos \alpha) + c$