The value of the definite integral `int_(-1)^(1)(1+x)^(1//2)(1-x)^(3//2)dx` equals

Options

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

Explanation

To solve the definite integral \( I = \int_{-1}^{1} (1+x)^{\frac{1}{2}} (1-x)^{\frac{3}{2}} \, dx \), we can follow these steps:

### Step 1: Define the integral
We start with the integral:
\[ I = \int_{-1}^{1} (1+x)^{\frac{1}{2}} (1-x)^{\frac{3}{2}} \, dx \]

### Step 2: Substitute \( x \) with \( -x \)
To utilize the properties of definite integrals, we can perform the substitution \( x \to -x \):
\[ I = \int_{-1}^{1} (1-x)^{\frac{1}{2}} (1+x)^{\frac{3}{2}} \, dx \]

### Step 3: Combine the two integrals
Now we have two expressions for \( I \):
1. \( I = \int_{-1}^{1} (1+x)^{\frac{1}{2}} (1-x)^{\frac{3}{2}} \, dx \)
2. \( I = \int_{-1}^{1} (1-x)^{\frac{1}{2}} (1+x)^{\frac{3}{2}} \, dx \)

Adding these two equations gives:
\[ 2I = \int_{-1}^{1} \left[ (1+x)^{\frac{1}{2}} (1-x)^{\frac{3}{2}} + (1-x)^{\frac{1}{2}} (1+x)^{\frac{3}{2}} \right] \, dx \]

### Step 4: Factor out common terms
We can factor out \( (1+x)^{\frac{1}{2}} (1-x)^{\frac{1}{2}} \):
\[ 2I = \int_{-1}^{1} (1+x)^{\frac{1}{2}} (1-x)^{\frac{1}{2}} \left[ (1-x) + (1+x) \right] \, dx \]
Simplifying the expression in the brackets:
\[ (1-x) + (1+x) = 2 \]
Thus, we have:
\[ 2I = \int_{-1}^{1} 2 (1+x)^{\frac{1}{2}} (1-x)^{\frac{1}{2}} \, dx \]
This simplifies to:
\[ I = \int_{-1}^{1} (1+x)^{\frac{1}{2}} (1-x)^{\frac{1}{2}} \, dx \]

### Step 5: Change the integral to a more manageable form
Using the identity \( (1+x)(1-x) = 1 - x^2 \), we can rewrite the integral:
\[ I = \int_{-1}^{1} \sqrt{(1-x^2)} \, dx \]

### Step 6: Recognize the integral as an even function
Since \( \sqrt{1-x^2} \) is an even function, we can simplify the integral:
\[ I = 2 \int_{0}^{1} \sqrt{1-x^2} \, dx \]

### Step 7: Evaluate the integral
The integral \( \int \sqrt{1-x^2} \, dx \) can be evaluated using the formula:
\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C \]
Here, \( a = 1 \):
\[ \int \sqrt{1-x^2} \, dx = \frac{x}{2} \sqrt{1-x^2} + \frac{1}{2} \sin^{-1}(x) + C \]
Evaluating from \( 0 \) to \( 1 \):
\[ \left[ \frac{x}{2} \sqrt{1-x^2} + \frac{1}{2} \sin^{-1}(x) \right]_{0}^{1} \]
At \( x = 1 \):
\[ \frac{1}{2} \cdot 0 + \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4} \]
At \( x = 0 \):
\[ \frac{0}{2} \cdot 1 + \frac{1}{2} \cdot 0 = 0 \]
Thus, the integral becomes:
\[ \int_{0}^{1} \sqrt{1-x^2} \, dx = \frac{\pi}{4} \]
So,
\[ I = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} \]

### Final Answer
The value of the definite integral is:
\[ \boxed{\frac{\pi}{2}} \]