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

Answer

To evaluate the integral

\[ \int \tan^{-1} \left( \sqrt{\frac{1-x}{1+x}} \right) dx, \]

we can follow these steps:

### Step 1: Substitution
Let \( x = \cos(2\theta) \). Then, we need to find \( dx \) in terms of \( d\theta \).

\[ dx = -2\sin(2\theta) d\theta \]

### Step 2: Rewrite the Integral
Now substitute \( x = \cos(2\theta) \) into the integral:

\[ \int \tan^{-1} \left( \sqrt{\frac{1 - \cos(2\theta)}{1 + \cos(2\theta)}} \right) (-2\sin(2\theta)) d\theta \]

### Step 3: Simplify the Argument of \( \tan^{-1} \)
Using the identities \( 1 - \cos(2\theta) = 2\sin^2(\theta) \) and \( 1 + \cos(2\theta) = 2\cos^2(\theta) \):

\[ \sqrt{\frac{1 - \cos(2\theta)}{1 + \cos(2\theta)}} = \sqrt{\frac{2\sin^2(\theta)}{2\cos^2(\theta)}} = \tan(\theta) \]

Thus, the integral becomes:

\[ \int \tan^{-1}(\tan(\theta)) (-2\sin(2\theta)) d\theta \]

### Step 4: Simplify \( \tan^{-1}(\tan(\theta)) \)
Since \( \tan^{-1}(\tan(\theta)) = \theta \) for \( \theta \) in the principal range, we have:

\[ \int \theta (-2\sin(2\theta)) d\theta \]

### Step 5: Integrate by Parts
Let \( u = \theta \) and \( dv = -2\sin(2\theta) d\theta \). Then \( du = d\theta \) and \( v = \int -2\sin(2\theta) d\theta = \cos(2\theta) \).

Using integration by parts:

\[ \int u \, dv = uv - \int v \, du \]

Thus:

\[ \int \theta (-2\sin(2\theta)) d\theta = \theta \cos(2\theta) - \int \cos(2\theta) d\theta \]

### Step 6: Evaluate the Remaining Integral
The integral of \( \cos(2\theta) \) is:

\[ \int \cos(2\theta) d\theta = \frac{1}{2}\sin(2\theta) \]

So we have:

\[ \theta \cos(2\theta) - \frac{1}{2}\sin(2\theta) + C \]

### Step 7: Back Substitute
Recall that \( \theta = \frac{1}{2} \cos^{-1}(x) \). Therefore:

\[ \theta \cos(2\theta) = \frac{1}{2} \cos^{-1}(x) \cdot \cos(\cos^{-1}(x)) = \frac{1}{2} \cos^{-1}(x) \cdot x \]

And \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) = 2\sqrt{1 - x^2} \cdot x \).

Putting it all together, we get:

\[ \frac{1}{2} x \cos^{-1}(x) - \frac{1}{2} \cdot 2\sqrt{1 - x^2} \cdot x + C \]

Thus, the final answer is:

\[ \frac{1}{2} x \cos^{-1}(x) - \sqrt{1 - x^2} \cdot x + C \]

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