Evaluate the following integrals:
`int frac{cosx-cos2x}{1-cosx}dx`

Answer

To evaluate the integral

\[ I = \int \frac{\cos x - \cos 2x}{1 - \cos x} \, dx, \]

we will follow a series of steps to simplify and solve the integral.

### Step 1: Rewrite the Integral
We start with the integral:

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

### Step 2: Use the Identity for Cosine
Recall the double angle identity for cosine:

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

Substituting this into the integral gives:

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

This simplifies to:

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

### Step 3: Simplify the Numerator
Now, we can rewrite the numerator:

\[ \cos x - 2 \cos^2 x + 1 = -2 \cos^2 x + \cos x + 1. \]

This can be rearranged as:

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

### Step 4: Factor the Numerator
We can factor the numerator:

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

Factoring \(2 \cos^2 x - \cos x - 1\) gives:

\[ (2 \cos x + 1)(\cos x - 1). \]

Thus, we have:

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

### Step 5: Cancel Out Terms
Notice that \(1 - \cos x = -(\cos x - 1)\), so we can cancel out \((\cos x - 1)\):

\[ I = \int -(2 \cos x + 1) \, dx. \]

### Step 6: Integrate
Now we can integrate:

\[ I = -\int (2 \cos x + 1) \, dx. \]

Calculating the integral:

\[ I = -\left(2 \sin x + x\right) + C, \]

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

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

\[ I = -2 \sin x - x + C. \]