The value of `int_(0)^(1) sqrt((1-x)/(1+x))dx` is

Options

A. `(pi)/(2)+1`
B. `(pi)/(2)-1`
C. -1
D. 1

Explanation

To solve the integral \( I = \int_{0}^{1} \sqrt{\frac{1-x}{1+x}} \, dx \), we will follow these steps:

### Step 1: Substitution
We will use the substitution \( x = \cos(2\theta) \). Then, we need to find \( dx \):
\[ dx = -2\sin(2\theta) \, d\theta \]

### Step 2: Change the limits of integration
When \( x = 0 \):
\[ \cos(2\theta) = 0 \implies 2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4} \]
When \( x = 1 \):
\[ \cos(2\theta) = 1 \implies 2\theta = 0 \implies \theta = 0 \]
Thus, the limits change from \( x = 0 \) to \( x = 1 \) into \( \theta = \frac{\pi}{4} \) to \( \theta = 0 \).

### Step 3: Rewrite the integral
Now, substituting \( x \) and \( dx \) into the integral:
\[ I = \int_{\frac{\pi}{4}}^{0} \sqrt{\frac{1 - \cos(2\theta)}{1 + \cos(2\theta)}} \cdot (-2\sin(2\theta)) \, d\theta \]
Reversing the limits gives:
\[ I = 2 \int_{0}^{\frac{\pi}{4}} \sqrt{\frac{1 - \cos(2\theta)}{1 + \cos(2\theta)}} \sin(2\theta) \, d\theta \]

### Step 4: Simplify the integrand
Using the trigonometric identities:
\[ 1 - \cos(2\theta) = 2\sin^2(\theta) \quad \text{and} \quad 1 + \cos(2\theta) = 2\cos^2(\theta) \]
Thus,
\[ \sqrt{\frac{1 - \cos(2\theta)}{1 + \cos(2\theta)}} = \sqrt{\frac{2\sin^2(\theta)}{2\cos^2(\theta)}} = \tan(\theta) \]

### Step 5: Substitute back into the integral
Now substituting this back into the integral:
\[ I = 2 \int_{0}^{\frac{\pi}{4}} \tan(\theta) \sin(2\theta) \, d\theta \]
Since \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \):
\[ I = 4 \int_{0}^{\frac{\pi}{4}} \sin^2(\theta) \cos(\theta) \, d\theta \]

### Step 6: Use the identity for sine
Using the identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \):
\[ I = 4 \int_{0}^{\frac{\pi}{4}} \frac{1 - \cos(2\theta)}{2} \cos(\theta) \, d\theta = 2 \int_{0}^{\frac{\pi}{4}} (1 - \cos(2\theta)) \cos(\theta) \, d\theta \]

### Step 7: Evaluate the integral
Now we can split the integral:
\[ I = 2 \left( \int_{0}^{\frac{\pi}{4}} \cos(\theta) \, d\theta - \int_{0}^{\frac{\pi}{4}} \cos(2\theta) \cos(\theta) \, d\theta \right) \]
The first integral:
\[ \int_{0}^{\frac{\pi}{4}} \cos(\theta) \, d\theta = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \]
The second integral can be solved using the product-to-sum identities:
\[ \int_{0}^{\frac{\pi}{4}} \cos(2\theta) \cos(\theta) \, d\theta = \frac{1}{2} \left( \int_{0}^{\frac{\pi}{4}} \cos(\theta) + \int_{0}^{\frac{\pi}{4}} \cos(3\theta) \right) \]
Calculating these gives:
\[ \frac{1}{2} \left( \frac{\sqrt{2}}{2} + 0 \right) = \frac{\sqrt{2}}{4} \]

### Step 8: Combine results
Now substituting back into our expression for \( I \):
\[ I = 2 \left( \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{4} \right) = 2 \cdot \frac{\sqrt{2}}{4} = \frac{\sqrt{2}}{2} \]

### Final Result
Thus, the value of the integral is:
\[ \boxed{\frac{\pi}{4} - \frac{1}{2}} \]