`int cos (2 cot^-1 sqrt((1-x)/(1+x)))dx`

Answer

To solve the integral \( \int \cos \left( 2 \cot^{-1} \sqrt{\frac{1-x}{1+x}} \right) dx \), we will follow these steps:

### Step 1: Simplify the Argument
We start with the expression inside the cosine function:
\[ \cot^{-1} \sqrt{\frac{1-x}{1+x}} \]

To simplify this, we can use the substitution \( x = \cos \theta \). This gives us:
\[ 1 - x = 1 - \cos \theta = 2 \sin^2 \frac{\theta}{2} \]
\[ 1 + x = 1 + \cos \theta = 2 \cos^2 \frac{\theta}{2} \]

Thus, we can rewrite the argument:
\[ \sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{2 \sin^2 \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}}} = \sqrt{\tan^2 \frac{\theta}{2}} = \tan \frac{\theta}{2} \]

### Step 2: Rewrite the Integral
Now, substituting back into the integral, we have:
\[ \int \cos \left( 2 \cot^{-1} \tan \frac{\theta}{2} \right) dx \]

Using the identity \( \cot^{-1} \tan \frac{\theta}{2} = \frac{\pi}{2} - \frac{\theta}{2} \), we can rewrite the cosine term:
\[ \cos \left( 2 \cot^{-1} \tan \frac{\theta}{2} \right) = \cos \left( \pi - \theta \right) = -\cos \theta \]

### Step 3: Substitute Back to x
Since we have \( \theta = \cos^{-1}(x) \), we can express \( -\cos \theta \) as:
\[ -\cos \theta = -x \]

### Step 4: Integrate
Now we can integrate:
\[ \int -x \, dx = -\frac{x^2}{2} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ -\frac{x^2}{2} + C \]

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