To solve the integral \( \int x \sqrt{\frac{1 - x}{1 + x}} \, dx \), we will follow a systematic approach.
### Step 1: Rationalization
We start by rationalizing the expression inside the integral. We rewrite the square root:
\[
\sqrt{\frac{1 - x}{1 + x}} = \frac{\sqrt{1 - x}}{\sqrt{1 + x}}
\]
Thus, the integral becomes:
\[
\int x \cdot \frac{\sqrt{1 - x}}{\sqrt{1 + x}} \, dx
\]
### Step 2: Substitution
Next, we will use the substitution \( x = \sin^2 \theta \). Then, we have:
\[
dx = 2 \sin \theta \cos \theta \, d\theta = \sin(2\theta) \, d\theta
\]
Now, substituting \( x \) in the integral:
\[
1 - x = 1 - \sin^2 \theta = \cos^2 \theta
\]
\[
1 + x = 1 + \sin^2 \theta = 1 + \sin^2 \theta
\]
The integral now transforms to:
\[
\int \sin^2 \theta \cdot \frac{\sqrt{\cos^2 \theta}}{\sqrt{1 + \sin^2 \theta}} \cdot \sin(2\theta) \, d\theta
\]
### Step 3: Simplifying the Integral
The integral simplifies to:
\[
\int \sin^2 \theta \cdot \cos \theta \cdot \sin(2\theta) \, d\theta
\]
Using \( \sin(2\theta) = 2 \sin \theta \cos \theta \):
\[
= \int \sin^2 \theta \cdot \cos \theta \cdot 2 \sin \theta \cos \theta \, d\theta = 2 \int \sin^3 \theta \cos^2 \theta \, d\theta
\]
### Step 4: Using Trigonometric Identity
We can use the identity \( \sin^2 \theta = 1 - \cos^2 \theta \):
\[
= 2 \int \sin^3 \theta (1 - \cos^2 \theta) \, d\theta
\]
### Step 5: Integration
Now, we can separate the integral:
\[
= 2 \left( \int \sin^3 \theta \, d\theta - \int \sin^3 \theta \cos^2 \theta \, d\theta \right)
\]
### Step 6: Solve Each Integral
The integral \( \int \sin^3 \theta \, d\theta \) can be solved using integration by parts or trigonometric identities.
### Step 7: Back Substitution
After solving the integral, we will substitute back \( \theta \) in terms of \( x \) using \( \theta = \sin^{-1}(\sqrt{x}) \).
### Final Result
After performing the back substitution and simplifying, we will arrive at the final expression for the integral.
