The value of `inte^(x)[(1+sinx)/(1+cosx)]dx` is equal to

Options

A. `(1)/(2)e^(x)sec.(x)/(2)+C`
B. `e^(x)sec.(x)/(2)+C`
C. `(1)/(2)e^(x)tan.(x)/(2)+C`
D. `e^(x)tan.(x)/(2)+C`

Explanation

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

### Step 1: Rewrite the integrand
We start by rewriting the integrand:
\[ \frac{1 + \sin x}{1 + \cos x} = \frac{1 + \sin x}{2 \cos^2 \frac{x}{2}} \quad \text{(since \(1 + \cos x = 2 \cos^2 \frac{x}{2}\))} \]
We can express \(1 + \sin x\) using the half-angle identity:
\[ 1 + \sin x = 1 + 2 \sin \frac{x}{2} \cos \frac{x}{2} = 1 + 2 \sin \frac{x}{2} \cos \frac{x}{2} \]

### Step 2: Substitute and simplify
Now, we can rewrite the integral:
\[ \int e^x \left( \frac{1}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} \right) \, dx \]
This simplifies to:
\[ \int e^x \left( \frac{1}{2 \cos^2 \frac{x}{2}} + \tan \frac{x}{2} \right) \, dx \]

### Step 3: Break the integral into two parts
We can split the integral into two parts:
\[ \int e^x \frac{1}{2 \cos^2 \frac{x}{2}} \, dx + \int e^x \tan \frac{x}{2} \, dx \]

### Step 4: Use integration by parts
For the second integral, we will use integration by parts. Let:
- \( u = \tan \frac{x}{2} \) and \( dv = e^x \, dx \)
Then,
- \( du = \frac{1}{2 \cos^2 \frac{x}{2}} \, dx \) and \( v = e^x \)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
This gives us:
\[ e^x \tan \frac{x}{2} - \int e^x \cdot \frac{1}{2 \cos^2 \frac{x}{2}} \, dx \]

### Step 5: Combine results
Now we can combine the results:
\[ \int e^x \frac{1 + \sin x}{1 + \cos x} \, dx = e^x \tan \frac{x}{2} + C \]

### Final Result
Thus, the value of the integral is:
\[ \int e^x \frac{1 + \sin x}{1 + \cos x} \, dx = e^x \tan \frac{x}{2} + C \]