`int(1)/(secx+tanx)dx=`

Options

A. `(1)/(sec)+c`
B. `secx+log(secx+tanx)+c`
C. `cosx+log(cosx-cotx)+c`
D. `log(secx)+c`

Answer

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

### Step 1: Rewrite the integrand
We start by rewriting \( \sec x \) and \( \tan x \) in terms of sine and cosine:
\[ \sec x = \frac{1}{\cos x}, \quad \tan x = \frac{\sin x}{\cos x} \]
Thus, we can express the integrand as:
\[ \sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x} \]
This allows us to rewrite the integral:
\[ \int \frac{1}{\sec x + \tan x} \, dx = \int \frac{\cos x}{1 + \sin x} \, dx \]

### Step 2: Use substitution
Next, we will use substitution to simplify the integral. Let:
\[ t = 1 + \sin x \]
Then, we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = \cos x \quad \Rightarrow \quad dt = \cos x \, dx \]
Now, we can substitute \( t \) and \( dt \) into the integral:
\[ \int \frac{\cos x}{1 + \sin x} \, dx = \int \frac{1}{t} \, dt \]

### Step 3: Integrate
The integral of \( \frac{1}{t} \) is:
\[ \int \frac{1}{t} \, dt = \ln |t| + C \]
Substituting back for \( t \):
\[ \ln |1 + \sin x| + C \]

### Final Answer
Thus, the final answer to the integral is:
\[ \int \frac{1}{\sec x + \tan x} \, dx = \ln |1 + \sin x| + C \]