`int sqrt((x)/( 1-x))dx` is equal to

Options

A. `sin^(-1) sqrt(x) + C`
B. `sin^(-1) (sqrt((x)(1-x)))+C`
C. `sin^(-1)(sqrt(x) - sqrt(x ( 1-x)))+C`
D. `sin^(-1) sqrt(x) - sqrt(x(1-x)) +C`

Answer

To solve the integral \( \int \sqrt{\frac{x}{1-x}} \, dx \), we will use a substitution method. Here’s the step-by-step solution:

### Step 1: Substitution
Let \( x = \sin^2 \theta \). Then, we differentiate to find \( dx \):
\[ dx = 2 \sin \theta \cos \theta \, d\theta \]

### Step 2: Substitute in the integral
Now, we substitute \( x \) and \( dx \) into the integral:
\[ \int \sqrt{\frac{\sin^2 \theta}{1 - \sin^2 \theta}} \cdot 2 \sin \theta \cos \theta \, d\theta \]
Since \( 1 - \sin^2 \theta = \cos^2 \theta \), we can simplify:
\[ \sqrt{\frac{\sin^2 \theta}{\cos^2 \theta}} = \frac{\sin \theta}{\cos \theta} \]
Thus, the integral becomes:
\[ \int \frac{\sin \theta}{\cos \theta} \cdot 2 \sin \theta \cos \theta \, d\theta = \int 2 \sin^2 \theta \, d\theta \]

### Step 3: Use a trigonometric identity
We can use the identity \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \):
\[ \int 2 \sin^2 \theta \, d\theta = \int (1 - \cos(2\theta)) \, d\theta \]

### Step 4: Integrate
Now we can integrate:
\[ \int (1 - \cos(2\theta)) \, d\theta = \theta - \frac{1}{2} \sin(2\theta) + C \]

### Step 5: Back-substitution
Now we need to substitute back in terms of \( x \). Since \( x = \sin^2 \theta \), we have:
\[ \theta = \sin^{-1}(\sqrt{x}) \]
For \( \sin(2\theta) \):
\[ \sin(2\theta) = 2 \sin \theta \cos \theta = 2 \sqrt{x} \sqrt{1-x} \]
Thus, substituting back:
\[ \int \sqrt{\frac{x}{1-x}} \, dx = \sin^{-1}(\sqrt{x}) - \sqrt{x(1-x)} + C \]

### Final Answer
The final result is:
\[ \int \sqrt{\frac{x}{1-x}} \, dx = \sin^{-1}(\sqrt{x}) - \sqrt{x(1-x)} + C \]