Solve the following integration
`int (dx)/(1+sin x)`

Answer

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

### Step-by-Step Solution:

1. **Rationalize the Denominator**:
We will multiply the numerator and the denominator by \( 1 - \sin x \):
\[ \int \frac{dx}{1 + \sin x} = \int \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} \, dx \]

2. **Simplify the Denominator**:
The denominator can be simplified using the identity \( 1 - \sin^2 x = \cos^2 x \):
\[ (1 + \sin x)(1 - \sin x) = 1 - \sin^2 x = \cos^2 x \]
Thus, the integral becomes:
\[ \int \frac{1 - \sin x}{\cos^2 x} \, dx \]

3. **Separate the Integral**:
We can separate the integral into two parts:
\[ \int \frac{1}{\cos^2 x} \, dx - \int \frac{\sin x}{\cos^2 x} \, dx \]
This can be rewritten as:
\[ \int \sec^2 x \, dx - \int \frac{\sin x}{\cos^2 x} \, dx \]

4. **Integrate Each Part**:
- The integral of \( \sec^2 x \) is:
\[ \int \sec^2 x \, dx = \tan x \]
- The integral of \( \frac{\sin x}{\cos^2 x} \) can be rewritten as \( \tan x \cdot \sec x \):
\[ \int \frac{\sin x}{\cos^2 x} \, dx = -\sec x \]
Thus, we have:
\[ \tan x + \sec x + C \]

5. **Final Result**:
Therefore, the final result of the integral is:
\[ \int \frac{dx}{1 + \sin x} = \tan x + \sec x + C \]