Evaluate the following integrals:
`int frac{cosx-cos^2x}{1-cosx}dx`

Answer

To evaluate the integral

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

we can follow these steps:

### Step 1: Simplify the integrand

First, we can factor out \(\cos x\) from the numerator:

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

So, we can rewrite the integral as:

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

### Step 2: Cancel common terms

Now, we can cancel the common term \(1 - \cos x\) in the numerator and the denominator (as long as \(1 - \cos x \neq 0\)):

\[ \int \cos x \, dx. \]

### Step 3: Integrate

The integral of \(\cos x\) is straightforward:

\[ \int \cos x \, dx = \sin x + C, \]

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

### Final Answer

Thus, the final result of the integral is:

\[ \sin x + C. \]

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