Area enclosed by the curve `(y-sin^(- 1)x)^2=x-x^2`, is (A) `pi/2` sq.units (B) `pi/4` sq.units (C) `pi/8` sq.units (D) none of these

Options

A. `pi//2` sq. units
B. `pi//4` sq. units
C. `pi//8` sq. units
D. None of these

Answer

To find the area enclosed by the curve \((y - \sin^{-1}x)^2 = x - x^2\), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ (y - \sin^{-1}x)^2 = x - x^2 \]
Taking the square root of both sides gives us two equations:
\[ y - \sin^{-1}x = \sqrt{x - x^2} \quad \text{and} \quad y - \sin^{-1}x = -\sqrt{x - x^2} \]
This leads to:
\[ y = \sin^{-1}x + \sqrt{x - x^2} \quad \text{(1)} \]
\[ y = \sin^{-1}x - \sqrt{x - x^2} \quad \text{(2)} \]

### Step 2: Finding Intersection Points
Next, we need to find the points where the curves intersect. The expression \(x - x^2\) can be factored:
\[ x - x^2 = x(1 - x) \]
This is zero when \(x = 0\) or \(x = 1\). Thus, the curves intersect at \(x = 0\) and \(x = 1\).

### Step 3: Setting Up the Area Integral
The area \(A\) between the two curves from \(x = 0\) to \(x = 1\) can be calculated using the integral:
\[ A = \int_0^1 \left( \left( \sin^{-1}x + \sqrt{x - x^2} \right) - \left( \sin^{-1}x - \sqrt{x - x^2} \right) \right) \, dx \]
This simplifies to:
\[ A = \int_0^1 2\sqrt{x - x^2} \, dx \]

### Step 4: Simplifying the Integral
We can rewrite \(x - x^2\) as:
\[ x - x^2 = x(1 - x) \]
Thus,
\[ \sqrt{x - x^2} = \sqrt{x(1 - x)} \]
So the area integral becomes:
\[ A = 2 \int_0^1 \sqrt{x(1 - x)} \, dx \]

### Step 5: Evaluating the Integral
The integral \(\int_0^1 \sqrt{x(1 - x)} \, dx\) can be evaluated using the beta function:
\[ \int_0^1 x^{1/2}(1-x)^{1/2} \, dx = B\left(\frac{3}{2}, \frac{3}{2}\right) = \frac{\Gamma\left(\frac{3}{2}\right) \Gamma\left(\frac{3}{2}\right)}{\Gamma(3)} = \frac{\left(\frac{\sqrt{\pi}}{2}\right)^2}{2} = \frac{\pi}{4} \]
Thus,
\[ \int_0^1 \sqrt{x(1 - x)} \, dx = \frac{\pi}{4} \]
Therefore, the area \(A\) becomes:
\[ A = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2} \]

### Conclusion
The area enclosed by the curve is:
\[ \frac{\pi}{2} \text{ square units} \]
So, the correct answer is (A) \(\frac{\pi}{2}\) sq.units.