`int(sinx)/((1+cosx))dx`

Answer

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

### Step 1: Identify the substitution
Let \( u = 1 + \cos x \).

### Step 2: Differentiate the substitution
Now, we differentiate \( u \):
\[ du = -\sin x \, dx \quad \Rightarrow \quad \sin x \, dx = -du \]

### Step 3: Substitute in the integral
Now we can substitute \( u \) and \( du \) into the integral:
\[ \int \frac{\sin x}{1 + \cos x} \, dx = \int \frac{\sin x}{u} \, dx = \int \frac{-du}{u} \]

### Step 4: Integrate
The integral \( \int \frac{-du}{u} \) can be solved as:
\[ -\ln |u| + C \]

### Step 5: Substitute back for \( u \)
Now we substitute back \( u = 1 + \cos x \):
\[ -\ln |1 + \cos x| + C \]

### Step 6: Simplify the expression
We can express this as:
\[ \ln \left( \frac{1}{1 + \cos x} \right) + C \]

### Final Answer
Thus, the final answer is:
\[ -\ln(1 + \cos x) + C \quad \text{or} \quad \ln \left( \frac{1}{1 + \cos x} \right) + C \]

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