Evaluate : `int (sin^2x)/(1+cosx) dx.`

Answer

To evaluate the integral \(\int \frac{\sin^2 x}{1 + \cos x} \, dx\), we can follow these steps:

### Step 1: Rewrite \(\sin^2 x\)
We know that \(\sin^2 x = 1 - \cos^2 x\). Thus, we can rewrite the integral as:
\[ \int \frac{1 - \cos^2 x}{1 + \cos x} \, dx \]

### Step 2: Split the Integral
Now, we can split the integral into two parts:
\[ \int \frac{1}{1 + \cos x} \, dx - \int \frac{\cos^2 x}{1 + \cos x} \, dx \]

### Step 3: Simplify the First Integral
For the first integral, we can use the identity \(1 + \cos x = 2 \cos^2 \left(\frac{x}{2}\right)\):
\[ \int \frac{1}{1 + \cos x} \, dx = \int \frac{1}{2 \cos^2 \left(\frac{x}{2}\right)} \, dx = \frac{1}{2} \int \sec^2 \left(\frac{x}{2}\right) \, dx \]
The integral of \(\sec^2 u\) is \(\tan u + C\), so:
\[ \frac{1}{2} \int \sec^2 \left(\frac{x}{2}\right) \, dx = \tan \left(\frac{x}{2}\right) + C_1 \]

### Step 4: Simplify the Second Integral
For the second integral, we can rewrite \(\cos^2 x\) using the identity \(\cos^2 x = \frac{1 + \cos 2x}{2}\):
\[ \int \frac{\cos^2 x}{1 + \cos x} \, dx = \int \frac{\frac{1 + \cos 2x}{2}}{1 + \cos x} \, dx = \frac{1}{2} \int \frac{1 + \cos 2x}{1 + \cos x} \, dx \]
This can be split into:
\[ \frac{1}{2} \int \frac{1}{1 + \cos x} \, dx + \frac{1}{2} \int \frac{\cos 2x}{1 + \cos x} \, dx \]
The first part is already calculated. The second part can be handled with substitution or further simplification.

### Step 5: Combine the Results
Combining the results from the first and second integrals, we have:
\[ \int \frac{\sin^2 x}{1 + \cos x} \, dx = \tan \left(\frac{x}{2}\right) - \left(\frac{1}{2} \tan \left(\frac{x}{2}\right) + C_2\right) + C \]
This simplifies to:
\[ \frac{1}{2} \tan \left(\frac{x}{2}\right) + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{\sin^2 x}{1 + \cos x} \, dx = \frac{1}{2} \tan \left(\frac{x}{2}\right) + C \]