`int(costheta-cos2theta)/(1-cos theta)d theta=`

Options

A. `theta+2sin theta+c`
B. `theta-2sin theta+c`
C. `2theta+2sin theta+c`
D. `2theta-2sin theta+c`

Answer

To solve the integral

\[ I = \int \frac{\cos \theta - \cos 2\theta}{1 - \cos \theta} \, d\theta, \]

we can follow these steps:

### Step 1: Use the identity for \(\cos 2\theta\)
We know that

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

Substituting this into the integral gives:

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

### Step 2: Simplify the numerator
Now, simplify the numerator:

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

So, the integral becomes:

\[ I = \int \frac{-2\cos^2 \theta + \cos \theta + 1}{1 - \cos \theta} \, d\theta. \]

### Step 3: Factor the numerator
Next, we can factor the numerator:

\[ -2\cos^2 \theta + \cos \theta + 1 = -2(\cos^2 \theta - \frac{1}{2}\cos \theta - \frac{1}{2}). \]

We can find the roots of the quadratic \(2x^2 - x - 1 = 0\) where \(x = \cos \theta\) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}. \]

Thus, the roots are:

\[ x = 1 \quad \text{and} \quad x = -\frac{1}{2}. \]

So, we can write:

\[ -2\cos^2 \theta + \cos \theta + 1 = -2(\cos \theta - 1)(\cos \theta + \frac{1}{2}). \]

### Step 4: Substitute back into the integral
Now, substituting this back into the integral gives:

\[ I = \int \frac{-2(\cos \theta - 1)(\cos \theta + \frac{1}{2})}{1 - \cos \theta} \, d\theta. \]

The \(1 - \cos \theta\) cancels out:

\[ I = \int -2(\cos \theta + \frac{1}{2}) \, d\theta. \]

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

\[ I = -2 \int \cos \theta \, d\theta - \int 1 \, d\theta. \]

### Step 6: Integrate
The integrals are:

\[ \int \cos \theta \, d\theta = \sin \theta, \]
and
\[ \int 1 \, d\theta = \theta. \]

Thus, we have:

\[ I = -2\sin \theta - \theta + C, \]

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

### Final Result
The final result is:

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

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