Evaluate the following integrals.
`int (1)/(1-cosx)dx`

Answer

To evaluate the integral \( I = \int \frac{1}{1 - \cos x} \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand using trigonometric identities
We know that \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \). Therefore, we can rewrite the integral as:
\[ I = \int \frac{1}{1 - \cos x} \, dx = \int \frac{1}{2 \sin^2\left(\frac{x}{2}\right)} \, dx \]

### Step 2: Simplify the integral
This can be simplified to:
\[ I = \frac{1}{2} \int \frac{1}{\sin^2\left(\frac{x}{2}\right)} \, dx \]
We know that \( \frac{1}{\sin^2 u} = \csc^2 u \), so we can write:
\[ I = \frac{1}{2} \int \csc^2\left(\frac{x}{2}\right) \, dx \]

### Step 3: Integrate using the known integral of cosecant squared
The integral of \( \csc^2 u \) is \( -\cot u + C \). Thus, we have:
\[ I = \frac{1}{2} \left(-\cot\left(\frac{x}{2}\right) + C\right) \]

### Step 4: Simplify the result
This simplifies to:
\[ I = -\frac{1}{2} \cot\left(\frac{x}{2}\right) + C \]

### Final Answer
Thus, the evaluated integral is:
\[ I = -\frac{1}{2} \cot\left(\frac{x}{2}\right) + C \]