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

Options

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

Answer

To solve the integral

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

we will follow these steps:

### Step 1: Rewrite the Numerator
Using the double angle formula for cosine, we can express \(\cos 2x\) and \(\cos 2\theta\):

\[ \cos 2x = 2\cos^2 x - 1 \quad \text{and} \quad \cos 2\theta = 2\cos^2 \theta - 1. \]

Thus, we can rewrite the numerator:

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

### Step 2: Factor the Numerator
Now, we can factor out the common term:

\[ 2(\cos^2 x - \cos^2 \theta). \]

### Step 3: Use the Difference of Squares
We can apply the difference of squares identity:

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

Thus, the integral becomes:

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

### Step 4: Simplify the Integral
We can cancel the \((\cos x - \cos \theta)\) terms in the numerator and denominator (assuming \(\cos x \neq \cos \theta\)):

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

### Step 5: Separate the Integral
Now we can separate the integral:

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

### Step 6: Integrate
The integral of \(\cos x\) is \(\sin x\), and since \(\cos \theta\) is a constant, its integral is \(\cos \theta \cdot x\):

\[ = 2 \left( \sin x + x \cos \theta \right) + C, \]

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

### Final Result
Thus, the final result of the integral is:

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