`int(x+sinx)/(1+cosx)\ dx` is equal to

Options

A. ` log|1+cosx|+C`
B. `log|x+sinx|+C`
C. `x-tan'(x)/(2)+C`
D. `x.tan'x/2 + C`

Explanation

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 \( \int \frac{x}{1 + \cos x} \, dx \)

To solve \( \int \frac{x}{1 + \cos x} \, dx \), we can use integration by parts. Let:

- \( u = x \) (which means \( du = dx \))
- \( dv = \frac{1}{1 + \cos x} \, dx \)

To find \( v \), we need to integrate \( \frac{1}{1 + \cos x} \). We can use the identity \( 1 + \cos x = 2 \cos^2(\frac{x}{2}) \):

\[ \int \frac{1}{1 + \cos x} \, dx = \int \frac{1}{2 \cos^2(\frac{x}{2})} \, dx = \frac{1}{2} \int \sec^2(\frac{x}{2}) \, dx \]

The integral of \( \sec^2(\frac{x}{2}) \) is \( 2 \tan(\frac{x}{2}) \):

\[ v = \tan\left(\frac{x}{2}\right) \]

Now applying integration by parts:

\[ \int u \, dv = uv - \int v \, du \]

This gives us:

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

### Step 2: Solve \( \int \tan\left(\frac{x}{2}\right) \, dx \)

The integral of \( \tan\left(\frac{x}{2}\right) \) can be solved using the identity:

\[ \tan\left(\frac{x}{2}\right) = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} \]

Using the substitution \( u = \frac{x}{2} \), we have \( dx = 2 \, du \):

\[ \int \tan\left(\frac{x}{2}\right) \, dx = 2 \int \tan(u) \, du = -2 \ln|\cos(u)| + C = -2 \ln|\cos\left(\frac{x}{2}\right)| + C \]

### Step 3: Combine Results

Now we can combine the results from Steps 1 and 2:

\[ \int \frac{x}{1 + \cos x} \, dx = x \tan\left(\frac{x}{2}\right) + 2 \ln|\cos\left(\frac{x}{2}\right)| + C \]

### Step 4: Solve \( \int \frac{\sin x}{1 + \cos x} \, dx \)

For the second integral, we can use the substitution \( u = 1 + \cos x \), which gives \( du = -\sin x \, dx \):

\[ \int \frac{\sin x}{1 + \cos x} \, dx = -\int \frac{1}{u} \, du = -\ln|u| + C = -\ln|1 + \cos x| + C \]

### Final Result

Combining both parts, we have:

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