The value of `int (cos x)/(sin x + cos x) dx ` is

Options

A. `(x^(2))/(2)+log|sinx+cosx|+C`
B. `(1)/(2)(x+log|sin x+cosx|)+C`
C. `2log|sinx+cosx|+x+C`
D. none of these

Answer

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

### Step 1: Rewrite the Integral
Let \( I = \int \frac{\cos x}{\sin x + \cos x} \, dx \).

### Step 2: Simplify the Numerator
We can express \( \cos x \) in terms of \( \sin x \) and \( \cos x \):
\[ \cos x = \frac{1}{2}(\sin x + \cos x) + \frac{1}{2}(\cos x - \sin x) \]
Thus, we can rewrite the integral as:
\[ I = \int \left( \frac{\frac{1}{2}(\sin x + \cos x) + \frac{1}{2}(\cos x - \sin x)}{\sin x + \cos x} \right) \, dx \]
This simplifies to:
\[ I = \int \left( \frac{1}{2} + \frac{\cos x - \sin x}{2(\sin x + \cos x)} \right) \, dx \]

### Step 3: Split the Integral
Now we can split the integral into two parts:
\[ I = \frac{1}{2} \int dx + \frac{1}{2} \int \frac{\cos x - \sin x}{\sin x + \cos x} \, dx \]

### Step 4: Solve the First Integral
The first integral is straightforward:
\[ \frac{1}{2} \int dx = \frac{1}{2} x \]

### Step 5: Change of Variables for the Second Integral
Let \( t = \sin x + \cos x \). Then, the derivative \( dt = (\cos x - \sin x) \, dx \).

### Step 6: Substitute in the Integral
Now we can rewrite the second integral:
\[ \frac{1}{2} \int \frac{dt}{t} \]
This integral evaluates to:
\[ \frac{1}{2} \ln |t| + C = \frac{1}{2} \ln |\sin x + \cos x| + C \]

### Step 7: Combine the Results
Combining both parts, we have:
\[ I = \frac{1}{2} x + \frac{1}{2} \ln |\sin x + \cos x| + C \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{1}{2} x + \frac{1}{2} \ln |\sin x + \cos x| + C \]