Area of region bounded by the curve `y=(16-x^(2))/(4)` and `y=sec^(-1)[-sin^(2)x]` (where [x] denotes the greatest ingeger function) is

Options

A. `(1)/(3)(4-pi)^(3//2)`
B. `8(4-pi)^(3//2)`
C. `(8)/(3)(4-pi)^(3//2)`
D. `(8)/(3)(4-pi)^(1//2)`

Explanation

To find the area of the region bounded by the curves \( y = \frac{16 - x^2}{4} \) and \( y = \sec^{-1}[-\sin^2 x] \) (where \([x]\) denotes the greatest integer function), we will follow these steps:

### Step 1: Simplify the second function
The function \( y = \sec^{-1}[-\sin^2 x] \) needs to be simplified. The range of \( \sin^2 x \) is from 0 to 1. Therefore, \( -\sin^2 x \) ranges from -1 to 0. The greatest integer function \([-\sin^2 x]\) will take the value -1 for all \( x \) in the interval where \( \sin^2 x \) is between 0 and 1. Thus:

\[ y = \sec^{-1}[-1] = \frac{\pi}{2} \]