Evaluate the following integrals:
`int (1)/(sec x + tan x)dx`

Answer

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

### Step 1: Rewrite the integrand
We start by rewriting the integrand in terms of sine and cosine. We know that:
\[ \sec x = \frac{1}{\cos x} \quad \text{and} \quad \tan x = \frac{\sin x}{\cos x} \]
Thus, we can express the integral as:
\[ \int \frac{1}{\sec x + \tan x} \, dx = \int \frac{1}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} \, dx \]
Now, we can combine the terms in the denominator:
\[ = \int \frac{1}{\frac{1 + \sin x}{\cos x}} \, dx = \int \frac{\cos x}{1 + \sin x} \, dx \]

### Step 2: Substitution
Next, we will use substitution. Let:
\[ t = \sin x \]
Then, the derivative of \( t \) with respect to \( x \) is:
\[ \frac{dt}{dx} = \cos x \quad \Rightarrow \quad dt = \cos x \, dx \]
Now we can rewrite the integral in terms of \( t \):
\[ \int \frac{\cos x}{1 + \sin x} \, dx = \int \frac{dt}{1 + t} \]

### Step 3: Integrate
Now we can integrate:
\[ \int \frac{dt}{1 + t} = \ln |1 + t| + C \]

### Step 4: Substitute back
Finally, we substitute back \( t = \sin x \):
\[ \ln |1 + \sin x| + C \]

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

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