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

Options

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

Answer

To solve the integral \( I = \int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx \), we can follow these steps:

### Step 1: Rationalize the Integral
We start by rewriting the integral:
\[ I = \int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx \]
To simplify this, we can multiply the integrand by \(\sqrt{1-x}\) and \(\sqrt{1-x}\) in the numerator and denominator:
\[ I = \int_0^1 \frac{(1-x)}{\sqrt{(1+x)(1-x)}} \, dx \]

### Step 2: Simplify the Expression
Using the identity \(a^2 - b^2 = (a-b)(a+b)\), we can express the denominator:
\[ I = \int_0^1 \frac{1-x}{\sqrt{1-x^2}} \, dx \]

### Step 3: Split the Integral
We can separate the integral into two parts:
\[ I = \int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx - \int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx \]
Let’s denote these two integrals as \(I_1\) and \(I_2\):
\[ I_1 = \int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx \]
\[ I_2 = \int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx \]

### Step 4: Evaluate \(I_1\)
The integral \(I_1\) can be evaluated using the known result:
\[ I_1 = \sin^{-1}(x) \Big|_0^1 = \sin^{-1}(1) - \sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2} \]

### Step 5: Evaluate \(I_2\)
For \(I_2\), we can use the substitution \(u = 1 - x^2\), which gives \(du = -2x \, dx\):
\[ I_2 = \int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx \]
Changing the limits accordingly, when \(x = 0\), \(u = 1\) and when \(x = 1\), \(u = 0\):
\[ I_2 = -\frac{1}{2} \int_1^0 u^{-1/2} \, du = \frac{1}{2} \int_0^1 u^{-1/2} \, du = \frac{1}{2} \left[ 2u^{1/2} \right]_0^1 = \frac{1}{2} \cdot 2 = 1 \]

### Step 6: Combine Results
Now we can combine the results of \(I_1\) and \(I_2\):
\[ I = I_1 - I_2 = \frac{\pi}{2} - 1 \]

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