`int (1-sinx )/ (1-cos x) dx=?`

Answer

To solve the integral \( \int \frac{1 - \sin x}{1 - \cos x} \, dx \), we can break it down into two parts. Let's denote the integral as \( I \).

### Step 1: Split the integral
We can rewrite the integral as:
\[ I = \int \left( \frac{1}{1 - \cos x} - \frac{\sin x}{1 - \cos x} \right) \, dx \]
This gives us two separate integrals:
\[ I = \int \frac{1}{1 - \cos x} \, dx - \int \frac{\sin x}{1 - \cos x} \, dx \]

**Hint for Step 1:** Consider breaking the integral into simpler parts for easier integration.

### Step 2: Solve the first integral
For the first integral \( \int \frac{1}{1 - \cos x} \, dx \), we can use the trigonometric identity:
\[ 1 - \cos x = 2 \sin^2 \left( \frac{x}{2} \right) \]
Thus,
\[ \int \frac{1}{1 - \cos x} \, dx = \int \frac{1}{2 \sin^2 \left( \frac{x}{2} \right)} \, dx = \frac{1}{2} \int \csc^2 \left( \frac{x}{2} \right) \, dx \]
The integral of \( \csc^2 \) is:
\[ \int \csc^2 u \, du = -\cot u \]
So,
\[ \frac{1}{2} \int \csc^2 \left( \frac{x}{2} \right) \, dx = -\frac{1}{2} \cot \left( \frac{x}{2} \right) \]

**Hint for Step 2:** Use trigonometric identities to simplify the integral.

### Step 3: Solve the second integral
Now, we focus on the second integral \( \int \frac{\sin x}{1 - \cos x} \, dx \). We can use the substitution \( u = \cos x \), which gives \( du = -\sin x \, dx \) or \( \sin x \, dx = -du \). The integral becomes:
\[ \int \frac{\sin x}{1 - \cos x} \, dx = -\int \frac{1}{1 - u} \, du \]
The integral of \( \frac{1}{1 - u} \) is:
\[ -\int \frac{1}{1 - u} \, du = -\log |1 - u| = -\log |1 - \cos x| \]

**Hint for Step 3:** Use substitution to transform the integral into a simpler form.

### Step 4: Combine the results
Now we can combine the results of the two integrals:
\[ I = -\frac{1}{2} \cot \left( \frac{x}{2} \right) + \log |1 - \cos x| + C \]

### Step 5: Simplify the expression
Using the identity \( 1 - \cos x = 2 \sin^2 \left( \frac{x}{2} \right) \), we can rewrite:
\[ \log |1 - \cos x| = \log |2 \sin^2 \left( \frac{x}{2} \right)| = \log 2 + 2 \log |\sin \left( \frac{x}{2} \right)| \]
Thus, the final expression for \( I \) becomes:
\[ I = -\frac{1}{2} \cot \left( \frac{x}{2} \right) + \log 2 + 2 \log |\sin \left( \frac{x}{2} \right)| + C \]

### Final Answer
The final answer for the integral is:
\[ I = -\frac{1}{2} \cot \left( \frac{x}{2} \right) + 2 \log |\sin \left( \frac{x}{2} \right)| + \log 2 + C \]