`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 simpler parts. Here’s the step-by-step solution:

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

### Step 2: Solve the First Integral
Now, we will integrate the first part:
\[ \int \frac{1}{1 - \cos x} \, dx \]
We can use the identity \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \):
\[ \int \frac{1}{2 \sin^2\left(\frac{x}{2}\right)} \, dx = \frac{1}{2} \int \csc^2\left(\frac{x}{2}\right) \, dx \]
The integral of \( \csc^2 u \) is \( -\cot u \):
\[ = -\cot\left(\frac{x}{2}\right) + C_1 \]

### Step 3: Solve the Second Integral
Next, we integrate the second part:
\[ -\int \frac{\sin x}{1 - \cos x} \, dx \]
Using the substitution \( t = 1 - \cos x \), we have \( dt = \sin x \, dx \). Thus, the integral becomes:
\[ -\int \frac{1}{t} \, dt = -\log |t| + C_2 = -\log |1 - \cos x| + C_2 \]

### Step 4: Combine the Results
Now, we combine both parts:
\[ \int \frac{1 - \sin x}{1 - \cos x} \, dx = -\cot\left(\frac{x}{2}\right) - \log |1 - \cos x| + C \]

### Final Answer
Thus, the final result is:
\[ \int \frac{1 - \sin x}{1 - \cos x} \, dx = -\cot\left(\frac{x}{2}\right) - \log |1 - \cos x| + C \]