Area bounded by `y = tan^(-1)x, y=cot^(-1) x` and y-axis is equal to

Options

A. ln `sqrt(2)` sq. unit
B. ln 4 sq. unit
C. ln 8 sq. unit
D. ln 2 sq. unit

Answer

To find the area bounded by the curves \( y = \tan^{-1} x \), \( y = \cot^{-1} x \), and the y-axis, we will follow these steps:

### Step 1: Find the intersection points of the curves
We need to find the point where \( y = \tan^{-1} x \) and \( y = \cot^{-1} x \) intersect. This occurs when:
\[ \tan^{-1} x = \cot^{-1} x \]
Using the identity \( \cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x \), we can set up the equation:
\[ \tan^{-1} x = \frac{\pi}{2} - \tan^{-1} x \]
Adding \( \tan^{-1} x \) to both sides gives:
\[ 2 \tan^{-1} x = \frac{\pi}{2} \]
Dividing both sides by 2, we find:
\[ \tan^{-1} x = \frac{\pi}{4} \]
Taking the tangent of both sides:
\[ x = 1 \]
Thus, the curves intersect at \( x = 1 \).

### Step 2: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) can be expressed as:
\[ A = \int_{0}^{1} (\cot^{-1} x - \tan^{-1} x) \, dx \]

### Step 3: Evaluate the integral
To evaluate the integral, we can use the fact that:
\[ \cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x \]
Thus, we can rewrite the integral:
\[ A = \int_{0}^{1} \left( \frac{\pi}{2} - \tan^{-1} x - \tan^{-1} x \right) \, dx = \int_{0}^{1} \left( \frac{\pi}{2} - 2\tan^{-1} x \right) \, dx \]
This can be split into two separate integrals:
\[ A = \int_{0}^{1} \frac{\pi}{2} \, dx - 2 \int_{0}^{1} \tan^{-1} x \, dx \]
Calculating the first integral:
\[ \int_{0}^{1} \frac{\pi}{2} \, dx = \frac{\pi}{2} \cdot 1 = \frac{\pi}{2} \]

### Step 4: Evaluate the integral of \( \tan^{-1} x \)
The integral of \( \tan^{-1} x \) can be computed using integration by parts:
Let \( u = \tan^{-1} x \) and \( dv = dx \), then \( du = \frac{1}{1+x^2} \, dx \) and \( v = x \).
Using integration by parts:
\[ \int \tan^{-1} x \, dx = x \tan^{-1} x - \int \frac{x}{1+x^2} \, dx \]
The second integral can be computed as:
\[ \int \frac{x}{1+x^2} \, dx = \frac{1}{2} \ln(1+x^2) \]
Thus,
\[ \int \tan^{-1} x \, dx = x \tan^{-1} x - \frac{1}{2} \ln(1+x^2) + C \]
Evaluating from 0 to 1:
\[ \left[ x \tan^{-1} x - \frac{1}{2} \ln(1+x^2) \right]_{0}^{1} = \left[ 1 \cdot \frac{\pi}{4} - \frac{1}{2} \ln(2) \right] - \left[ 0 - 0 \right] = \frac{\pi}{4} - \frac{1}{2} \ln(2) \]

### Step 5: Combine results to find the area
Putting it all together:
\[ A = \frac{\pi}{2} - 2 \left( \frac{\pi}{4} - \frac{1}{2} \ln(2) \right) = \frac{\pi}{2} - \frac{\pi}{2} + \ln(2) = \ln(2) \]

### Final Result:
Thus, the area bounded by the curves is:
\[ \boxed{\ln(2)} \]