`int tan^-1(sqrt((1-x)/(1+x)) dx`

Answer

To solve the integral \( I = \int \tan^{-1} \left( \sqrt{\frac{1-x}{1+x}} \right) dx \), we will follow these steps:

### Step 1: Substitution
Let \( x = \cos(\theta) \). Then, we have:
\[ dx = -\sin(\theta) d\theta \]

### Step 2: Simplifying the Expression
Substituting \( x = \cos(\theta) \) into the integral:
\[ I = \int \tan^{-1} \left( \sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}} \right)(- \sin(\theta) d\theta) \]

Using the identity \( 1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right) \) and \( 1 + \cos(\theta) = 2\cos^2\left(\frac{\theta}{2}\right) \), we get:
\[ \sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}} = \sqrt{\frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\cos^2\left(\frac{\theta}{2}\right)}} = \tan\left(\frac{\theta}{2}\right) \]

Thus, the integral becomes:
\[ I = -\int \tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right) \sin(\theta) d\theta \]

### Step 3: Simplifying the Integral
Since \( \tan^{-1}(\tan(x)) = x \) when \( x \) is in the principal range, we have:
\[ I = -\int \frac{\theta}{2} \sin(\theta) d\theta \]

### Step 4: Integration by Parts
Using integration by parts, let:
- \( u = \frac{\theta}{2} \) then \( du = \frac{1}{2} d\theta \)
- \( dv = \sin(\theta) d\theta \) then \( v = -\cos(\theta) \)

Applying integration by parts:
\[ I = -\left( -\frac{\theta}{2} \cos(\theta) + \int \cos(\theta) \cdot \frac{1}{2} d\theta \right) \]
\[ = \frac{\theta}{2} \cos(\theta) - \frac{1}{2} \sin(\theta) \]

### Step 5: Back Substitution
Now, we substitute back \( \theta = \cos^{-1}(x) \):
\[ I = \frac{\cos^{-1}(x)}{2} \cdot \sqrt{1 - x^2} - \frac{1}{2} x + C \]

### Final Answer
Thus, the integral evaluates to:
\[ I = \frac{1}{2} \cos^{-1}(x) \sqrt{1 - x^2} - \frac{x}{2} + C \]