If `I=underset (0)overset(1) int cos{ 2 "cot"^(-1)sqrt((1-x)/(1+x))}dx` then

Options

A. `I gt (1)/(2)`
B. `I = -(1)/(2)`
C. `0 lt I lt (1)/(2)`
D. None of these

Answer

To solve the integral \( I = \int_0^1 \cos\left(2 \cot^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right) dx \), we will follow these steps:

### Step 1: Simplify the expression inside the cosine
We start by rewriting the argument of the cosine function. We know that:
\[ \cot^{-1}(y) = \frac{\pi}{2} - \tan^{-1}(y) \]
Thus, we can express \( \cot^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right) \) in terms of \( \tan^{-1} \).

### Step 2: Substitute \( x = \cos(2\theta) \)
Let \( x = \cos(2\theta) \). Then:
\[ 1 - x = 1 - \cos(2\theta) = 2\sin^2(\theta) \]
\[ 1 + x = 1 + \cos(2\theta) = 2\cos^2(\theta) \]
So, we have:
\[ \sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{2\sin^2(\theta)}{2\cos^2(\theta)}} = \tan(\theta) \]
Thus, we can rewrite the integral as:
\[ I = \int_0^1 \cos(2 \cot^{-1}(\tan(\theta))) dx \]

### Step 3: Use the cotangent identity
Using the identity \( \cot^{-1}(\tan(\theta)) = \frac{\pi}{2} - \theta \), we have:
\[ I = \int_0^1 \cos(2(\frac{\pi}{2} - \theta)) dx = \int_0^1 \cos(\pi - 2\theta) dx \]
Using the property \( \cos(\pi - x) = -\cos(x) \):
\[ I = -\int_0^1 \cos(2\theta) dx \]

### Step 4: Change of variable
Now, we need to express \( dx \) in terms of \( d\theta \). From \( x = \cos(2\theta) \), we differentiate:
\[ dx = -2\sin(2\theta) d\theta \]
The limits change as follows:
- When \( x = 0 \), \( \theta = \frac{\pi}{4} \)
- When \( x = 1 \), \( \theta = 0 \)

Thus, the integral becomes:
\[ I = -\int_{\frac{\pi}{4}}^0 \cos(2\theta)(-2\sin(2\theta)) d\theta = 2\int_0^{\frac{\pi}{4}} \cos(2\theta)\sin(2\theta) d\theta \]

### Step 5: Use the double angle identity
Using the identity \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \), we have:
\[ \cos(2\theta)\sin(2\theta) = \frac{1}{2} \sin(4\theta) \]
Thus, the integral simplifies to:
\[ I = \int_0^{\frac{\pi}{4}} \sin(4\theta) d\theta \]

### Step 6: Evaluate the integral
The integral of \( \sin(4\theta) \) is:
\[ -\frac{1}{4}\cos(4\theta) \bigg|_0^{\frac{\pi}{4}} = -\frac{1}{4} \left( \cos(\pi) - \cos(0) \right) = -\frac{1}{4}(-1 - 1) = \frac{1}{2} \]

### Final Result
Thus, we conclude that:
\[ I = -\frac{1}{2} \]