Question

The area of the region bounded by the curve $y=x^2$ and $y={\sec }^{-1}[-{\sin }^{2}x],$ (where $[.]$ denotes the greatest integer function), is $\frac{4\pi \sqrt{\pi }}{n}$ sq. unit. Find the value of $n$.

Answer 1

$[-{\sin }^{2}x]=0$ or $-1$ but ${\sec }^{-1}\left(0\right)$ is not defined.
$\Rightarrow {\sec }^{-1}[-{\sin }^{2}x]={\sec }^{-1}(-1)=\pi$
The required area$={\int }_{-\sqrt{\pi }}^{\sqrt{\pi }}(\pi -x^2)dx$
$={[\pi x-\frac{x^3}{3}]}_{-\sqrt{\pi }}^{\sqrt{\pi }}$
$=\frac{4}{3}\pi \sqrt{\pi }$ (on simplification)

Hence, the value of $n$ is $3$.