To evaluate the integral \( I = \int_{-1}^{1} (1+x)^{\frac{1}{2}} (1-x)^{\frac{3}{2}} \, dx \), we will use the property of definite integrals that states:
\[
\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx
\]
### Step 1: Set up the integral
We start with the integral:
\[
I = \int_{-1}^{1} (1+x)^{\frac{1}{2}} (1-x)^{\frac{3}{2}} \, dx
\]
### Step 2: Apply the property of definite integrals
Using the property mentioned, we replace \( x \) with \( -x \):
\[
I = \int_{-1}^{1} (1 - x)^{\frac{1}{2}} (1 + x)^{\frac{3}{2}} \, dx
\]
### Step 3: Write down the two expressions for \( I \)
Now we have two expressions for \( I \):
1. \( I = \int_{-1}^{1} (1+x)^{\frac{1}{2}} (1-x)^{\frac{3}{2}} \, dx \) (Equation 1)
2. \( I = \int_{-1}^{1} (1-x)^{\frac{1}{2}} (1+x)^{\frac{3}{2}} \, dx \) (Equation 2)
### Step 4: Add the two equations
Adding both 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 5: Factor out common terms
We can factor out \( (1+x)^{\frac{1}{2}} (1-x)^{\frac{1}{2}} \):
\[
2I = \int_{-1}^{1} (1+x)^{\frac{1}{2}} (1-x)^{\frac{1}{2}} \left[ (1-x) + (1+x) \right] \, dx
\]
### Step 6: Simplify the expression inside the integral
The expression simplifies to:
\[
(1-x) + (1+x) = 2
\]
Thus, we have:
\[
2I = \int_{-1}^{1} (1+x)^{\frac{1}{2}} (1-x)^{\frac{1}{2}} \cdot 2 \, dx
\]
### Step 7: Simplify further
This gives:
\[
2I = 2 \int_{-1}^{1} (1+x)^{\frac{1}{2}} (1-x)^{\frac{1}{2}} \, dx
\]
### Step 8: Divide by 2
Dividing both sides by 2, we get:
\[
I = \int_{-1}^{1} (1+x)^{\frac{1}{2}} (1-x)^{\frac{1}{2}} \, dx
\]
### Step 9: Change of variables
We can use the substitution \( x = \sin(\theta) \), which gives \( dx = \cos(\theta) d\theta \) and changes the limits from \( -1 \) to \( 1 \) (corresponding to \( \theta = -\frac{\pi}{2} \) to \( \frac{\pi}{2} \)):
\[
I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1+\sin(\theta))^{\frac{1}{2}} (1-\sin(\theta))^{\frac{1}{2}} \cos(\theta) \, d\theta
\]
### Step 10: Evaluate the integral
Using the identity \( (1+\sin(\theta))(1-\sin(\theta)) = \cos^2(\theta) \):
\[
I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(\theta) \cos(\theta) \, d\theta
\]
This simplifies to:
\[
I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^3(\theta) \, d\theta
\]
### Step 11: Solve the integral
The integral of \( \cos^3(\theta) \) can be computed using the reduction formula or known results:
\[
I = \frac{3\pi}{8}
\]
### Final Result
Thus, the value of \( I \) is:
\[
I = \frac{\pi}{2}
\]
