` int _(0)^(1) x^(3//2) sqrt(1-x) dx ` is equal to

Options

A. `pi/6`
B. `pi/9`
C. `pi/12`
D. `pi/16`

Explanation

To solve the integral \( \int_0^1 x^{\frac{3}{2}} \sqrt{1-x} \, dx \), we can use the substitution method. Here are the steps to find the solution:

### Step 1: Substitution
Let \( x = \sin^2 \theta \). Then, we have:
\[ dx = 2 \sin \theta \cos \theta \, d\theta = \sin(2\theta) \, d\theta \]

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

### Step 3: Substitute into the integral
Now substitute \( x \) and \( dx \) into the integral:
\[ \int_0^1 x^{\frac{3}{2}} \sqrt{1-x} \, dx = \int_0^{\frac{\pi}{2}} (\sin^2 \theta)^{\frac{3}{2}} \sqrt{1 - \sin^2 \theta} \cdot \sin(2\theta) \, d\theta \]
This simplifies to:
\[ = \int_0^{\frac{\pi}{2}} \sin^3 \theta \cdot \cos \theta \cdot \sin(2\theta) \, d\theta \]
Since \( \sin(2\theta) = 2 \sin \theta \cos \theta \), we can rewrite the integral as:
\[ = 2 \int_0^{\frac{\pi}{2}} \sin^4 \theta \cos^2 \theta \, d\theta \]

### Step 4: Simplifying the integral
Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \), we can express the integral as:
\[ = 2 \int_0^{\frac{\pi}{2}} \sin^4 \theta (1 - \sin^2 \theta) \, d\theta \]
This expands to:
\[ = 2 \left( \int_0^{\frac{\pi}{2}} \sin^4 \theta \, d\theta - \int_0^{\frac{\pi}{2}} \sin^6 \theta \, d\theta \right) \]

### Step 5: Evaluating the integrals
Using the Beta function, we know:
\[ \int_0^{\frac{\pi}{2}} \sin^{2n} \theta \, d\theta = \frac{1}{2} B(n + \frac{1}{2}, \frac{1}{2}) = \frac{1}{2} \cdot \frac{\Gamma(n + \frac{1}{2}) \Gamma(\frac{1}{2})}{\Gamma(n + 1)} \]
Calculating for \( n = 2 \) and \( n = 3 \):
\[ \int_0^{\frac{\pi}{2}} \sin^4 \theta \, d\theta = \frac{3}{8} \cdot \frac{\pi}{2} = \frac{3\pi}{16} \]
\[ \int_0^{\frac{\pi}{2}} \sin^6 \theta \, d\theta = \frac{5}{16} \cdot \frac{\pi}{2} = \frac{5\pi}{32} \]

### Step 6: Combine the results
Substituting these values back:
\[ = 2 \left( \frac{3\pi}{16} - \frac{5\pi}{32} \right) \]
Finding a common denominator:
\[ = 2 \left( \frac{6\pi}{32} - \frac{5\pi}{32} \right) = 2 \cdot \frac{\pi}{32} = \frac{\pi}{16} \]

### Final Answer
Thus, the value of the integral \( \int_0^1 x^{\frac{3}{2}} \sqrt{1-x} \, dx \) is:
\[ \frac{\pi}{16} \]