`int (x)/( 1+cos x) dx=?`

Answer

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

### Step 1: Simplify the Denominator
We know that:
\[ 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right) \]
Thus, we can rewrite the integral as:
\[ \int \frac{x}{1 + \cos x} \, dx = \int \frac{x}{2 \cos^2\left(\frac{x}{2}\right)} \, dx = \frac{1}{2} \int \frac{x}{\cos^2\left(\frac{x}{2}\right)} \, dx \]

### Step 2: Use Integration by Parts
Let:
- \( u = x \) (which implies \( du = dx \))
- \( dv = \sec^2\left(\frac{x}{2}\right) \, dx \)

Now we need to find \( v \):
\[ v = \int \sec^2\left(\frac{x}{2}\right) \, dx \]
The integral of \( \sec^2\left(\frac{x}{2}\right) \) is:
\[ v = 2 \tan\left(\frac{x}{2}\right) \]

### Step 3: Apply Integration by Parts Formula
Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
we have:
\[ \int \frac{x}{\cos^2\left(\frac{x}{2}\right)} \, dx = x \cdot 2 \tan\left(\frac{x}{2}\right) - \int 2 \tan\left(\frac{x}{2}\right) \, dx \]

### Step 4: Solve the Remaining Integral
Now we need to solve \( \int 2 \tan\left(\frac{x}{2}\right) \, dx \):
\[ \int 2 \tan\left(\frac{x}{2}\right) \, dx = -4 \log\left|\cos\left(\frac{x}{2}\right)\right| + C \]

### Step 5: Combine the Results
Putting everything together, we have:
\[ \int \frac{x}{\cos^2\left(\frac{x}{2}\right)} \, dx = 2x \tan\left(\frac{x}{2}\right) + 4 \log\left|\cos\left(\frac{x}{2}\right)\right| + C \]

### Step 6: Final Result
Therefore, the final result for the integral is:
\[ \int \frac{x}{1 + \cos x} \, dx = \frac{1}{2} \left( 2x \tan\left(\frac{x}{2}\right) + 4 \log\left|\cos\left(\frac{x}{2}\right)\right| \right) + C \]
This simplifies to:
\[ x \tan\left(\frac{x}{2}\right) + 2 \log\left|\cos\left(\frac{x}{2}\right)\right| + C \]