`int_(0)^(1)x^(3)sqrt(1-x^(2))dx=`

Options

A. `(1)/(15)`
B. `(22)/(15)`
C. `(4)/(15)`
D. `(2)/(15)`

Answer

To solve the integral \( \int_{0}^{1} x^{3} \sqrt{1 - x^{2}} \, dx \), we will follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ \int_{0}^{1} x^{3} \sqrt{1 - x^{2}} \, dx \]
We can rewrite \( x^{3} \) as \( x \cdot x^{2} \) and factor out \( \sqrt{1 - x^{2}} \):
\[ = \int_{0}^{1} x \cdot x^{2} \sqrt{1 - x^{2}} \, dx \]

### Step 2: Use substitution
Let us use the substitution \( t^{2} = 1 - x^{2} \). Then, differentiating both sides gives:
\[ -2x \, dx = 2t \, dt \quad \Rightarrow \quad x \, dx = -t \, dt \]
We also need to change the limits of integration. When \( x = 0 \), \( t = 1 \) and when \( x = 1 \), \( t = 0 \). Thus, the limits change from \( 0 \) to \( 1 \) into \( 1 \) to \( 0 \).

### Step 3: Substitute in the integral
Substituting \( x^{2} = 1 - t^{2} \) and \( x \, dx = -t \, dt \):
\[ = \int_{1}^{0} (1 - t^{2}) \sqrt{t^{2}} (-t) \, dt \]
This simplifies to:
\[ = \int_{1}^{0} (1 - t^{2}) t (-t) \, dt = \int_{1}^{0} -t^{2} (1 - t^{2}) \, dt \]
Reversing the limits gives:
\[ = \int_{0}^{1} t^{2} (1 - t^{2}) \, dt \]

### Step 4: Expand the integrand
Now, we expand the integrand:
\[ = \int_{0}^{1} (t^{2} - t^{4}) \, dt \]

### Step 5: Integrate term by term
Now we can integrate each term:
\[ = \left[ \frac{t^{3}}{3} - \frac{t^{5}}{5} \right]_{0}^{1} \]
Calculating this gives:
\[ = \left( \frac{1^{3}}{3} - \frac{1^{5}}{5} \right) - \left( \frac{0^{3}}{3} - \frac{0^{5}}{5} \right) \]
\[ = \frac{1}{3} - \frac{1}{5} \]

### Step 6: Find a common denominator and simplify
To combine these fractions, we find a common denominator:
\[ = \frac{5}{15} - \frac{3}{15} = \frac{2}{15} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{1} x^{3} \sqrt{1 - x^{2}} \, dx = \frac{2}{15} \]

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