`int (x+sin x)/(1-cos x)dx`

Answer

To solve the integral \( \int \frac{x + \sin x}{1 - \cos x} \, dx \), we can break it down into two separate integrals:

\[ \int \frac{x + \sin x}{1 - \cos x} \, dx = \int \frac{x}{1 - \cos x} \, dx + \int \frac{\sin x}{1 - \cos x} \, dx \]

### Step 1: Solve \( \int \frac{x}{1 - \cos x} \, dx \)

Using the identity \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \), we can rewrite the integral:

\[ \int \frac{x}{1 - \cos x} \, dx = \int \frac{x}{2 \sin^2\left(\frac{x}{2}\right)} \, dx = \frac{1}{2} \int \frac{x}{\sin^2\left(\frac{x}{2}\right)} \, dx \]

### Step 2: Solve \( \int \frac{\sin x}{1 - \cos x} \, dx \)

Using the same identity:

\[ \int \frac{\sin x}{1 - \cos x} \, dx = \int \frac{\sin x}{2 \sin^2\left(\frac{x}{2}\right)} \, dx = \frac{1}{2} \int \frac{\sin x}{\sin^2\left(\frac{x}{2}\right)} \, dx \]

### Step 3: Combine the two integrals

Now we can combine the results:

\[ \int \frac{x + \sin x}{1 - \cos x} \, dx = \frac{1}{2} \int \frac{x}{\sin^2\left(\frac{x}{2}\right)} \, dx + \frac{1}{2} \int \frac{\sin x}{\sin^2\left(\frac{x}{2}\right)} \, dx \]

### Step 4: Evaluate the integrals

1. **For \( \int \frac{x}{\sin^2\left(\frac{x}{2}\right)} \, dx \)**, we can use integration by parts or substitution.
2. **For \( \int \frac{\sin x}{\sin^2\left(\frac{x}{2}\right)} \, dx \)**, we can simplify using trigonometric identities.

### Step 5: Final Result

After evaluating both integrals, we can combine the results and add the constant of integration \( C \):

\[ \int \frac{x + \sin x}{1 - \cos x} \, dx = \text{(result from first integral)} + \text{(result from second integral)} + C \]