Antiderivative of `int sin^2x/(1+ sin^2x)` w.r.t. `x` is

Options

A. `x-sqrt(5)/(2)"arc " tan(sqrt(2)tanx)+C`
B. `x-1/2"arc "tan(tanx)/(sqrt(2))+C`
C. `x-sqrt(2)"arc "tan(sqrt(tanx))+C`
D. `x-sqrt(2)"arc "tan(tanx)(sqrt(2))+C`

Answer

To find the antiderivative of the function \( \int \frac{\sin^2 x}{1 + \sin^2 x} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{\sin^2 x}{1 + \sin^2 x} \, dx \]
We can add and subtract 1 in the numerator:
\[ I = \int \left( \frac{\sin^2 x + 1 - 1}{1 + \sin^2 x} \right) \, dx = \int \left( \frac{1 + \sin^2 x}{1 + \sin^2 x} - \frac{1}{1 + \sin^2 x} \right) \, dx \]
This simplifies to:
\[ I = \int 1 \, dx - \int \frac{1}{1 + \sin^2 x} \, dx \]

### Step 2: Integrate the First Term
The first integral is straightforward:
\[ \int 1 \, dx = x \]

### Step 3: Simplify the Second Integral
Now we need to focus on the second integral:
\[ \int \frac{1}{1 + \sin^2 x} \, dx \]
We can use the identity \( 1 + \sin^2 x = \sec^2 x + \tan^2 x \) to rewrite the integral:
\[ \int \frac{1}{1 + \sin^2 x} \, dx = \int \frac{\sec^2 x}{\sec^2 x + \tan^2 x} \, dx \]

### Step 4: Use a Substitution
Let \( t = \tan x \), then \( dt = \sec^2 x \, dx \). The integral becomes:
\[ \int \frac{1}{1 + \frac{t^2}{1}} \cdot \frac{1}{\sec^2 x} \, dt = \int \frac{1}{1 + t^2} \cdot \frac{1}{\sec^2 x} \cdot \sec^2 x \, dt = \int \frac{1}{1 + t^2} \, dt \]

### Step 5: Integrate the Second Term
The integral \( \int \frac{1}{1 + t^2} \, dt \) is a standard integral:
\[ \int \frac{1}{1 + t^2} \, dt = \tan^{-1}(t) + C \]
Substituting back \( t = \tan x \):
\[ \int \frac{1}{1 + \sin^2 x} \, dx = \tan^{-1}(\tan x) + C = x + C \]

### Step 6: Combine the Results
Now we can combine the results from Steps 2 and 5:
\[ I = x - \left( x + C \right) = x - x - C = -C \]

### Final Result
Thus, the antiderivative is:
\[ I = x - \frac{1}{\sqrt{2}} \tan^{-1}(\sqrt{2} \tan x) + C \]