Evaluate: `int(cos2x-cos2theta)/(cosx-costheta)dx`

Options

A. `2(sinx+xcostheta) + C`
B. `2(sinx - xcos theta) + C`
C. `2(sinx+2xcostheta) + C`
D. `2(sinx-2x cos theta)+C`

Explanation

To evaluate the integral

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

we can follow these steps:

### Step 1: Rewrite the numerator using the double angle formula
We know that

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

Thus, we can rewrite the integral as:

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

### Step 2: Simplify the numerator
This simplifies to:

\[ I = \int \frac{2 \cos^2 x - 2 \cos^2 \theta}{\cos x - \cos \theta} \, dx. \]

Factoring out the 2 gives us:

\[ I = 2 \int \frac{\cos^2 x - \cos^2 \theta}{\cos x - \cos \theta} \, dx. \]

### Step 3: Use the difference of squares
The expression \(\cos^2 x - \cos^2 \theta\) can be factored using the difference of squares:

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

Substituting this back into the integral gives:

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

### Step 4: Cancel out the common terms
The \((\cos x - \cos \theta)\) terms cancel out:

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

### Step 5: Split the integral
Now we can split the integral:

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

### Step 6: Evaluate the integrals
The integral of \(\cos x\) is \(\sin x\), and since \(\cos \theta\) is a constant, we have:

\[ \int \cos \theta \, dx = \cos \theta \cdot x. \]

Putting it all together, we get:

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

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

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

\[ I = 2 \sin x + 2 x \cos \theta + C. \]

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