Evaluate `int(1)/(1+sinx)dx`.

Answer

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

### Step 1: Rationalize the Denominator
We start with the integral:
\[ I = \int \frac{1}{1+\sin x} \, dx \]
To simplify this, we multiply the numerator and the denominator by \( 1 - \sin x \):
\[ I = \int \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} \, dx \]

### Step 2: Simplify the Denominator
The denominator can be simplified using the difference of squares:
\[ (1 + \sin x)(1 - \sin x) = 1 - \sin^2 x \]
Using the Pythagorean identity \( \cos^2 x = 1 - \sin^2 x \), we rewrite the integral:
\[ I = \int \frac{1 - \sin x}{\cos^2 x} \, dx \]

### Step 3: Separate the Integral
Now we can separate the integral into two parts:
\[ I = \int \frac{1}{\cos^2 x} \, dx - \int \frac{\sin x}{\cos^2 x} \, dx \]
This gives us:
\[ I = \int \sec^2 x \, dx - \int \tan x \sec x \, dx \]

### Step 4: Integrate Each Part
Now we can integrate each part:
1. The integral of \( \sec^2 x \) is \( \tan x \):
\[ \int \sec^2 x \, dx = \tan x \]
2. The integral of \( \tan x \sec x \) is \( \sec x \):
\[ \int \tan x \sec x \, dx = \sec x \]

Putting it all together, we have:
\[ I = \tan x - \sec x + C \]

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