The area bounded by the curves $y =\sin^{–1} |\sin x|$ and $y = (\sin^{–1} |\sin x|)^2, 0 ≤ x ≤ 2π$ is

$(\frac{π^3}{6}-\frac{π^2}{2}+\frac{4}{3})$ sq unit

$∵\sin^{–1} |\sin x|=\left\{\begin{matrix}x,&0≤x≤\frac{π}{2}\\π-x,&\frac{π}{2}≤x≤π\\-π+x,&π≤x≤\frac{3π}{2}\\2π-x,&\frac{3π}{2}≤x≤2π\end{matrix}\right.$

$∴(\sin^{–1} |\sin x|)^2=\left\{\begin{matrix}x^2,&0≤x≤\frac{π}{2}\\(π-x)^2,&\frac{π}{2}≤x≤π\\(x-π)^2,&π≤x≤\frac{3π}{2}\\(2π-x)^2,&\frac{3π}{2}≤x≤2π\end{matrix}\right.$

∴ Required area

$=4\left\{\int_0^1(x-x^2)dx+\int_1^{π/2}(x^2-x)dx\right\}$

$=4\left\{\left[\left(\frac{x^2}{2}-\frac{x^3}{3}\right)\right]_0^1+\left[\left(\frac{x^3}{3}-\frac{x^2}{2}\right)\right]_1^{π/2}\right\}$

$=4\left\{1-\frac{2}{3}+\frac{π^3}{24}-\frac{π^2}{8}\right\}=\left\{\frac{π^3}{6}-\frac{π^2}{2}+\frac{4}{3}\right\}$ Sq unit