Evaluate: `int cos {2 tan^(-1) sqrt((1-x)/(1+x))}dx`

Answer

To evaluate the integral \( \int \cos(2 \tan^{-1}(\sqrt{\frac{1-x}{1+x}})) \, dx \), we can use a substitution method. Here’s the step-by-step solution:

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

### Step 2: Rewrite the Integral
Now, we need to rewrite the integral in terms of \( \theta \):
\[ \int \cos(2 \tan^{-1}(\sqrt{\frac{1 - \cos(2\theta)}{1 + \cos(2\theta)}})) \, dx \]
Using the identity \( \tan^{-1}(t) = \frac{1}{2} \tan^{-1}(\frac{2t}{1-t^2}) \), we can simplify \( \tan^{-1}(\sqrt{\frac{1 - \cos(2\theta)}{1 + \cos(2\theta)}}) \).

### Step 3: Simplify the Argument of Cosine
Using the trigonometric identities:
- \( 1 - \cos(2\theta) = 2\sin^2(\theta) \)
- \( 1 + \cos(2\theta) = 2\cos^2(\theta) \)

This gives us:
\[ \sqrt{\frac{1 - \cos(2\theta)}{1 + \cos(2\theta)}} = \sqrt{\frac{2\sin^2(\theta)}{2\cos^2(\theta)}} = \tan(\theta) \]
Thus, we have:
\[ \cos(2 \tan^{-1}(\tan(\theta))) = \cos(2\theta) \]

### Step 4: Substitute Back into the Integral
Now substituting back, we have:
\[ \int \cos(2\theta) (-2 \sin(2\theta)) \, d\theta \]
This simplifies to:
\[ -2 \int \cos(2\theta) \sin(2\theta) \, d\theta \]

### Step 5: Use Trigonometric Identity
Using the identity \( \sin(2\theta) \cos(2\theta) = \frac{1}{2} \sin(4\theta) \):
\[ -2 \int \frac{1}{2} \sin(4\theta) \, d\theta = -\int \sin(4\theta) \, d\theta \]

### Step 6: Integrate
The integral of \( \sin(4\theta) \) is:
\[ -\int \sin(4\theta) \, d\theta = \frac{1}{4} \cos(4\theta) + C \]

### Step 7: Substitute Back to x
Recall that \( x = \cos(2\theta) \). We need to express \( \cos(4\theta) \) in terms of \( x \):
\[ \cos(4\theta) = 2\cos^2(2\theta) - 1 = 2x^2 - 1 \]
Thus, we have:
\[ \frac{1}{4} (2x^2 - 1) + C = \frac{x^2}{2} - \frac{1}{4} + C \]

### Final Answer
The final result is:
\[ \int \cos(2 \tan^{-1}(\sqrt{\frac{1-x}{1+x}})) \, dx = \frac{x^2}{2} + C' \]
where \( C' \) is a constant of integration.