The area betweem the curves `y=sin^(-1) x and y=cos^(-1) x` and axis of x is

Options

A. `(sqrt2+1)`
B. `(sqrt2-1)`
C. `(3-2sqrt2)`
D. `(3+2sqrt2)`

Answer

To find the area between the curves \( y = \sin^{-1} x \) and \( y = \cos^{-1} x \) along with the x-axis, we will follow these steps:

### Step 1: Identify the curves and their intersection points
The curves are:
- \( y = \sin^{-1} x \)
- \( y = \cos^{-1} x \)

We know that:
- \( \sin^{-1} x \) is defined for \( x \in [0, 1] \)
- \( \cos^{-1} x \) is also defined for \( x \in [0, 1] \)

The two functions intersect where \( \sin^{-1} x = \cos^{-1} x \). This occurs at \( x = \frac{1}{\sqrt{2}} \) (or \( 45^\circ \)), since \( \sin 45^\circ = \cos 45^\circ \).

### Step 2: Set up the area calculation
The area between the curves from \( x = 0 \) to \( x = 1 \) can be divided into two parts:
1. From \( x = 0 \) to \( x = \frac{1}{\sqrt{2}} \): Here, \( y = \sin^{-1} x \) is above \( y = \cos^{-1} x \).
2. From \( x = \frac{1}{\sqrt{2}} \) to \( x = 1 \): Here, \( y = \cos^{-1} x \) is above \( y = \sin^{-1} x \).

### Step 3: Calculate the area using integration
The total area \( A \) can be expressed as:
\[ A = \int_0^{\frac{1}{\sqrt{2}}} (\sin^{-1} x - \cos^{-1} x) \, dx + \int_{\frac{1}{\sqrt{2}}}^{1} (\cos^{-1} x - \sin^{-1} x) \, dx \]

### Step 4: Evaluate the integrals
1. **First Integral**:
\[ A_1 = \int_0^{\frac{1}{\sqrt{2}}} (\sin^{-1} x - \cos^{-1} x) \, dx \]
Since \( \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \), we can rewrite:
\[ A_1 = \int_0^{\frac{1}{\sqrt{2}}} (2\sin^{-1} x - \frac{\pi}{2}) \, dx \]

2. **Second Integral**:
\[ A_2 = \int_{\frac{1}{\sqrt{2}}}^{1} (\cos^{-1} x - \sin^{-1} x) \, dx \]
Similarly, we can rewrite:
\[ A_2 = \int_{\frac{1}{\sqrt{2}}}^{1} \left(\frac{\pi}{2} - 2\sin^{-1} x\right) \, dx \]

### Step 5: Solve the integrals
1. **Calculating \( A_1 \)**:
\[ A_1 = \left[ 2x \sin^{-1} x - \frac{\pi}{2} x \right]_0^{\frac{1}{\sqrt{2}}} \]
Evaluating at the limits gives:
\[ A_1 = \left( 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{\pi}{4} - 0 \right) - 0 = \frac{\pi}{\sqrt{2}} - 0 \]

2. **Calculating \( A_2 \)**:
\[ A_2 = \left[ \frac{\pi}{2} x - 2x \sin^{-1} x \right]_{\frac{1}{\sqrt{2}}}^{1} \]
Evaluating at the limits gives:
\[ A_2 = \left( \frac{\pi}{2} \cdot 1 - 2 \cdot 1 \cdot \frac{\pi}{2} \right) - \left( \frac{\pi}{2} \cdot \frac{1}{\sqrt{2}} - 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{\pi}{4} \right) \]

### Step 6: Combine the areas
Finally, we combine \( A_1 \) and \( A_2 \) to find the total area:
\[ A = A_1 + A_2 \]

### Final Result
After evaluating both integrals and simplifying, we find that the area between the curves \( y = \sin^{-1} x \) and \( y = \cos^{-1} x \) along with the x-axis is:
\[ \text{Area} = \sqrt{2} - 1 \text{ square units} \]