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

Options

A. 2
B. 1
C. `-1`
D. 0

Answer

To solve the integral

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

we can use a substitution method.

### Step 1: Substitution
Let \( t = \sin^{-1}(x) \). Then, we differentiate to find \( dx \):
\[ x = \sin(t) \quad \Rightarrow \quad dx = \cos(t) \, dt. \]
Also, we need to change the limits of integration. When \( x = -1 \), \( t = -\frac{\pi}{2} \) and when \( x = 1 \), \( t = \frac{\pi}{2} \).

### Step 2: Change the integral
Substituting \( x \) and \( dx \) into the integral gives:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(t) \cdot t \cdot \frac{\cos(t)}{\sqrt{1 - \sin^2(t)}} \, dt. \]
Since \( \sqrt{1 - \sin^2(t)} = \cos(t) \), we can simplify this to:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} t \cdot \sin(t) \, dt. \]

### Step 3: Integration by parts
We will use integration by parts, where we let:
- \( u = t \) and \( dv = \sin(t) \, dt \).
Then, we differentiate and integrate:
- \( du = dt \) and \( v = -\cos(t) \).

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ I = \left[ -t \cos(t) \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} -\cos(t) \, dt. \]

### Step 4: Evaluate the boundary term
Calculating the boundary term:
\[ -t \cos(t) \bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \left(-\frac{\pi}{2} \cdot \cos\left(\frac{\pi}{2}\right)\right) - \left(-\left(-\frac{\pi}{2}\right) \cdot \cos\left(-\frac{\pi}{2}\right)\right) = 0 - 0 = 0. \]

### Step 5: Evaluate the remaining integral
Now we need to evaluate:
\[ I = 0 + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(t) \, dt. \]
The integral of \( \cos(t) \) is:
\[ \int \cos(t) \, dt = \sin(t) \bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin\left(-\frac{\pi}{2}\right) = 1 - (-1) = 2. \]

### Final Result
Thus, the value of the integral is:
\[ I = 2. \]