`int(sinx)/(1+sin x) dx`

Answer

To solve the integral \( \int \frac{\sin x}{1 + \sin x} \, dx \), we can use the method of rationalization. Here’s a step-by-step solution:

### Step 1: Rationalize the Integrand
We start by multiplying the integrand by a form of 1, specifically \( \frac{1 - \sin x}{1 - \sin x} \):
\[ \int \frac{\sin x}{1 + \sin x} \, dx = \int \frac{\sin x (1 - \sin x)}{(1 + \sin x)(1 - \sin x)} \, dx \]

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

### Step 3: Expand the Numerator
Now, we expand the numerator:
\[ \sin x (1 - \sin x) = \sin x - \sin^2 x \]
So the integral becomes:
\[ \int \left( \frac{\sin x}{\cos^2 x} - \frac{\sin^2 x}{\cos^2 x} \right) \, dx \]

### Step 4: Split the Integral
We can split the integral into two separate integrals:
\[ \int \frac{\sin x}{\cos^2 x} \, dx - \int \frac{\sin^2 x}{\cos^2 x} \, dx \]

### Step 5: Substitute for Each Integral
1. For the first integral, we know that:
\[ \frac{\sin x}{\cos^2 x} = \tan x \sec x \]
Therefore:
\[ \int \frac{\sin x}{\cos^2 x} \, dx = \int \tan x \sec x \, dx \]
The integral of \( \tan x \sec x \) is \( \sec x \).

2. For the second integral:
\[ \frac{\sin^2 x}{\cos^2 x} = \tan^2 x \]
Thus:
\[ \int \frac{\sin^2 x}{\cos^2 x} \, dx = \int \tan^2 x \, dx \]
We know that \( \tan^2 x = \sec^2 x - 1 \), so:
\[ \int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx = \tan x - x \]

### Step 6: Combine the Results
Now we combine the results from both integrals:
\[ \sec x - (\tan x - x) + C = \sec x - \tan x + x + C \]

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