`int(sinx)/((1+sinx))dx=?`

Options

A. `tanx+secx+C`
B. `tanx-secx+C`
C. `(1)/(2)"tan"(x)/(2)+C`
D. none of these

Explanation

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

### Step 1: Simplify the integrand
We start by multiplying and dividing the integrand by \( 1 - \sin x \):
\[ \int \frac{\sin x}{1 + \sin x} \, dx = \int \frac{\sin x (1 - \sin x)}{(1 + \sin x)(1 - \sin x)} \, dx \]
This simplifies to:
\[ \int \frac{\sin x (1 - \sin x)}{\cos^2 x} \, dx \]
because \( (1 + \sin x)(1 - \sin x) = 1 - \sin^2 x = \cos^2 x \).

### Step 2: Split the integral
Now we can split the integral into two parts:
\[ \int \frac{\sin x}{\cos^2 x} \, dx - \int \frac{\sin^2 x}{\cos^2 x} \, dx \]
This can be rewritten as:
\[ \int \tan x \sec^2 x \, dx - \int \tan^2 x \, dx \]

### Step 3: Integrate the first term
The integral \( \int \tan x \sec^2 x \, dx \) can be solved using the substitution \( u = \tan x \), which gives us:
\[ \int \tan x \sec^2 x \, dx = \frac{1}{2} \tan^2 x + C_1 \]

### Step 4: Integrate the second term
For the integral \( \int \tan^2 x \, dx \), we can use the identity \( \tan^2 x = \sec^2 x - 1 \):
\[ \int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx = \int \sec^2 x \, dx - \int 1 \, dx \]
This results in:
\[ \tan x - x + C_2 \]

### Step 5: Combine the results
Now we combine the results from both integrals:
\[ \int \frac{\sin x}{1 + \sin x} \, dx = \frac{1}{2} \tan^2 x - (\tan x - x) + C \]
Simplifying this gives:
\[ \int \frac{\sin x}{1 + \sin x} \, dx = \frac{1}{2} \tan^2 x - \tan x + x + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{\sin x}{1 + \sin x} \, dx = \frac{1}{2} \tan^2 x - \tan x + x + C \]