Evaluate the following integrals:
`int (1-cos x)/(sin x) dx`

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \frac{1 - \cos x}{\sin x} \, dx \]
We can separate the fraction:
\[ \int \left( \frac{1}{\sin x} - \frac{\cos x}{\sin x} \right) \, dx = \int \csc x \, dx - \int \cot x \, dx \]

### Step 2: Integrate Each Term
Now we will integrate each term separately.

1. The integral of \( \csc x \):
\[ \int \csc x \, dx = -\ln |\csc x + \cot x| + C_1 \]

2. The integral of \( \cot x \):
\[ \int \cot x \, dx = \ln |\sin x| + C_2 \]

### Step 3: Combine the Results
Now we combine the results of the integrals:
\[ \int \frac{1 - \cos x}{\sin x} \, dx = -\ln |\csc x + \cot x| - \ln |\sin x| + C \]

Using the property of logarithms \( \ln a - \ln b = \ln \frac{a}{b} \):
\[ = -\ln \left( \frac{|\csc x + \cot x|}{|\sin x|} \right) + C \]

### Step 4: Simplify the Expression
We can simplify further:
\[ = -\ln \left( \frac{1 + \cos x}{\sin x} \right) + C \]
where \( \csc x = \frac{1}{\sin x} \) and \( \cot x = \frac{\cos x}{\sin x} \).

### Final Result
Thus, the final answer is:
\[ \int \frac{1 - \cos x}{\sin x} \, dx = -\ln \left( \frac{1 + \cos x}{\sin x} \right) + C \]