`int_(-1)^(1)sqrt((1-x)/(1+x))dx=`

Options

A. `(pi)/4`
B. `(pi)/3`
C. `(pi)/2`
D. `pi`

Answer

To solve the integral \( I = \int_{-1}^{1} \sqrt{\frac{1-x}{1+x}} \, dx \), we can use the property of definite integrals and symmetry. Here’s a step-by-step solution:

### Step 1: Define the Integral
Let
\[ I = \int_{-1}^{1} \sqrt{\frac{1-x}{1+x}} \, dx \]

### Step 2: Use the Property of Definite Integrals
We can use the property that states:
\[ \int_{-a}^{a} f(x) \, dx = \int_{0}^{a} f(x) \, dx + \int_{0}^{a} f(-x) \, dx \]
In our case, we will find \( f(-x) \):
\[ f(-x) = \sqrt{\frac{1+x}{1-x}} \]

### Step 3: Rewrite the Integral
Now we can rewrite the integral \( I \) as:
\[ I = \int_{0}^{1} \sqrt{\frac{1-x}{1+x}} \, dx + \int_{0}^{1} \sqrt{\frac{1+x}{1-x}} \, dx \]

### Step 4: Combine the Integrals
Now, we can combine these two integrals:
\[ I = \int_{0}^{1} \left( \sqrt{\frac{1-x}{1+x}} + \sqrt{\frac{1+x}{1-x}} \right) \, dx \]

### Step 5: Simplify the Expression
To simplify the expression inside the integral, we can find a common denominator:
\[ \sqrt{\frac{1-x}{1+x}} + \sqrt{\frac{1+x}{1-x}} = \frac{\sqrt{(1-x)^2} + \sqrt{(1+x)^2}}{\sqrt{(1+x)(1-x)}} \]
This simplifies to:
\[ \frac{(1-x) + (1+x)}{\sqrt{1-x^2}} = \frac{2}{\sqrt{1-x^2}} \]

### Step 6: Substitute Back into the Integral
Now substituting back, we have:
\[ I = \int_{0}^{1} \frac{2}{\sqrt{1-x^2}} \, dx \]

### Step 7: Evaluate the Integral
The integral \( \int \frac{1}{\sqrt{1-x^2}} \, dx \) is a standard integral that evaluates to \( \sin^{-1}(x) \):
\[ I = 2 \left[ \sin^{-1}(x) \right]_{0}^{1} = 2 \left( \frac{\pi}{2} - 0 \right) = \pi \]

### Final Result
Thus, the value of the integral is:
\[ I = \pi \]