`int ( cos 2x - cos 2 alpha)/( cos x - cos alpha) dx ` is equal to

Options

A. `2 ( sin x + x cos alpha) + C`
B. `2 ( sin x - x cos alpha ) + C`
C. ` 2 ( sin x + 2 x cos alpha) = C`
D. ` 2 ( sin x - 2 x cos alpha ) + C`

Answer

To solve the integral

\[ \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} \, dx, \]

we can start by applying the double angle formula for cosine. The double angle formula states that

\[ \cos 2\theta = 2\cos^2 \theta - 1. \]

### Step 1: Rewrite the integrand using the double angle formula

Using this formula, we can rewrite \(\cos 2x\) and \(\cos 2\alpha\):

\[ \cos 2x = 2\cos^2 x - 1, \]
\[ \cos 2\alpha = 2\cos^2 \alpha - 1. \]

Substituting these into the integral gives:

\[ \int \frac{(2\cos^2 x - 1) - (2\cos^2 \alpha - 1)}{\cos x - \cos \alpha} \, dx. \]

### Step 2: Simplify the expression

Now simplify the numerator:

\[ (2\cos^2 x - 1) - (2\cos^2 \alpha - 1) = 2\cos^2 x - 2\cos^2 \alpha. \]

This can be factored:

\[ = 2(\cos^2 x - \cos^2 \alpha). \]

Thus, the integral becomes:

\[ \int \frac{2(\cos^2 x - \cos^2 \alpha)}{\cos x - \cos \alpha} \, dx. \]

### Step 3: Factor the numerator

Using the identity \(a^2 - b^2 = (a - b)(a + b)\), we can factor \(\cos^2 x - \cos^2 \alpha\):

\[ \cos^2 x - \cos^2 \alpha = (\cos x - \cos \alpha)(\cos x + \cos \alpha). \]

Substituting this back into the integral gives:

\[ \int \frac{2(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} \, dx. \]

### Step 4: Cancel the common terms

The \((\cos x - \cos \alpha)\) terms in the numerator and denominator cancel out (as long as \(\cos x \neq \cos \alpha\)):

\[ \int 2(\cos x + \cos \alpha) \, dx. \]

### Step 5: Integrate

Now we can integrate:

\[ = 2\int (\cos x + \cos \alpha) \, dx = 2\left(\int \cos x \, dx + \int \cos \alpha \, dx\right). \]

The integral of \(\cos x\) is \(\sin x\) and the integral of \(\cos \alpha\) is \(\cos \alpha \cdot x\):

\[ = 2(\sin x + x \cos \alpha) + C, \]

where \(C\) is the constant of integration.

### Final Answer

Thus, the final result is:

\[ \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} \, dx = 2\sin x + 2x \cos \alpha + C. \]