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

Options

A. `"tan"(x)/(2)+2log|"cos"(x)/(2)|+C`
B. `"-tan"(x)/(2)+2log|"cos"(x)/(2)|+C`
C. `"tan"(x)/(2)-2log|"cos"(x)/(2)|+C`
D. none of these

Explanation

To solve the integral \( \int \frac{1 + \sin x}{1 + \cos x} \, dx \), we can break it down into simpler parts. Here’s a step-by-step solution:

### Step 1: Split the Integral
We can separate the integral into two parts:
\[ \int \frac{1 + \sin x}{1 + \cos x} \, dx = \int \frac{1}{1 + \cos x} \, dx + \int \frac{\sin x}{1 + \cos x} \, dx \]

### Step 2: Solve the First Integral
For the first integral \( \int \frac{1}{1 + \cos x} \, dx \), 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 \( \sec^2 \) is:
\[ \frac{1}{2} \cdot 2 \tan \left(\frac{x}{2}\right) = \tan \left(\frac{x}{2}\right) \]

### Step 3: Solve the Second Integral
Now, we tackle the second integral \( \int \frac{\sin x}{1 + \cos x} \, dx \). We can use the substitution \( u = 1 + \cos x \), which gives \( du = -\sin x \, dx \) or \( \sin x \, dx = -du \):
\[ \int \frac{\sin x}{1 + \cos x} \, dx = -\int \frac{1}{u} \, du = -\ln |u| + C = -\ln |1 + \cos x| + C \]

### Step 4: Combine the Results
Now we combine the results from both integrals:
\[ \int \frac{1 + \sin x}{1 + \cos x} \, dx = \tan \left(\frac{x}{2}\right) - \ln |1 + \cos x| + C \]

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