`int_(-1)^(1) (x^(2) sin^(-1) [x])/(sqrt""(1-x^(2)))dx=`

Options

A. 1
B. 2
C. 0
D. none

Answer

To solve the integral

\[ I = \int_{-1}^{1} \frac{x^2 \sin^{-1}(x)}{\sqrt{1 - x^2}} \, dx, \]

we can follow these steps:

### Step 1: Substitution
Let's use the substitution \( t = \sin^{-1}(x) \). Then, we have:

\[ x = \sin(t) \quad \text{and} \quad dx = \cos(t) \, dt = \sqrt{1 - \sin^2(t)} \, dt = \sqrt{1 - x^2} \, dt. \]

### Step 2: Change of Limits
When \( x = -1 \), \( t = \sin^{-1}(-1) = -\frac{\pi}{2} \).
When \( x = 1 \), \( t = \sin^{-1}(1) = \frac{\pi}{2} \).

Thus, the limits of integration change from \([-1, 1]\) to \([- \frac{\pi}{2}, \frac{\pi}{2}]\).

### Step 3: Rewrite the Integral
Substituting \( x \) and \( dx \) into the integral, we get:

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2(t) \cdot t}{\sqrt{1 - \sin^2(t)}} \cdot \cos(t) \, dt. \]

Since \( \sqrt{1 - \sin^2(t)} = \cos(t) \), we can simplify the integral:

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2(t) \cdot t \, dt. \]

### Step 4: Check the Function's Symmetry
We need to check if \( f(-t) = -f(t) \) for the function \( f(t) = \sin^2(t) \cdot t \):

\[ f(-t) = \sin^2(-t) \cdot (-t) = \sin^2(t) \cdot (-t) = -f(t). \]

Since \( f(-t) = -f(t) \), the function is odd.

### Step 5: Evaluate the Integral
For an odd function integrated over a symmetric interval around zero, the integral evaluates to zero:

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(t) \, dt = 0. \]

### Final Result
Thus, the value of the integral is:

\[ I = 0. \]