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

Options

A. `Igt(1)/(2)`
B. `I=-(1)/(2)`
C. `0lt I lt (1)/(2)`
D. `I=(1)/(2)`

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 Argument of Cosine
We start with the expression inside the cosine function. We know that:
\[ 2 \cot^{-1}(x) = 2 \tan^{-1}\left(\frac{1}{x}\right) \]
Thus, we can rewrite:
\[ \cot^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{\frac{1-x}{1+x}}}\right) = \tan^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right) \]

### Step 2: Use the Double Angle Identity for Cosine
Using the identity for cosine of double angles:
\[ \cos(2\theta) = 1 - 2\sin^2(\theta) \]
We can express our integral as:
\[ I = \int_0^1 \left(1 - 2\sin^2\left(\tan^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right)\right)\right) dx \]

### Step 3: Find the Value of \(\sin\) and \(\cos\)
From the definition of \(\tan\):
\[ \tan(\theta) = \sqrt{\frac{1+x}{1-x}} \implies \sin(\theta) = \frac{\sqrt{1+x}}{\sqrt{(1+x)+(1-x)}} = \frac{\sqrt{1+x}}{\sqrt{2}} \quad \text{and} \quad \cos(\theta) = \frac{\sqrt{1-x}}{\sqrt{2}} \]

### Step 4: Substitute Back into the Integral
Now we substitute back into our integral:
\[ I = \int_0^1 \left(1 - 2\left(\frac{\sqrt{1+x}}{\sqrt{2}}\right)^2\right) dx = \int_0^1 \left(1 - \frac{2(1+x)}{2}\right) dx = \int_0^1 (1 - (1+x)) dx = \int_0^1 -x \, dx \]

### Step 5: Evaluate the Integral
Now we evaluate:
\[ I = -\int_0^1 x \, dx = -\left[\frac{x^2}{2}\right]_0^1 = -\left(\frac{1^2}{2} - 0\right) = -\frac{1}{2} \]

### Final Result
Thus, the value of the integral is:
\[ I = -\frac{1}{2} \]