Find the Area Bounded by $y = \frac{2}{π} [|\cos^{-1}(\sin x)| - |\sin^{-1}(\cos x)|]$ and $x$-axis between $ \frac{3π}{2}≤x≤2π $.

Question

Find the area bounded by $\displaystyle y = \dfrac{2}{π} \left[ \space \left|\cos^{-1}(\sin x)\right| - \left|\sin^{-1}(\cos x)\right| \space \right]$ and $x$-axis between $ \dfrac{3\pi}{2}≤x≤2\pi $. Here, $\big[ \space $.$ \space \big]$ means the greatest integer function.

My approach was using $\cos^{-1}x = \frac{π}{2} - \sin^{-1}x$ and the same for $\sin^{-1}x$.

$\displaystyle \therefore y = \dfrac{2}{π} \left[ \space \left|\frac{π}{2} - \sin^{-1}(\sin x)\right| - \left|\frac{π}{2} - \cos^{-1}(\cos x)\right| \space \right] $

$ \sin^{-1}(\sin x) = x$ only when $\displaystyle x \in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right] $

$\cos^{-1}(\cos x) = x$ only when $x \in \left[0, \pi \right]$

Since $ \sin(x-2π) = \sin x$ and $ \cos(2π-x) = \cos x $,

$\displaystyle y = \dfrac{2}{\pi} \left[ \space \left|\dfrac{\pi}{2} - (x-2\pi)\right| - \left|\frac{\pi}{2} - (2\pi-x)\right| \space \right] \iff y = \dfrac{2}{\pi} \left[ \space \left|\dfrac{5\pi}{2}-x\right| - \left|-\dfrac{3\pi}{2}-x\right| \space \right] $

Then, $\displaystyle \left|\dfrac{5\pi}{2}-x\right| > 0$ and $\displaystyle -\left|-\dfrac{3\pi}{2}-x\right| < 0 \implies y = \dfrac{2}{\pi}[\pi-2x]$

After this I do not know how to calculate the area or draw the graph of this function.

Answer 1

$$y = \frac{2}{\pi}\left[\left|\frac{5\pi}2-x - (x-\frac{3\pi}2)\right|\right]=\frac2{\pi}[4\pi-2x]$$

We have $$\frac{3\pi}{2}\le x \le 2\pi,$$ $$3\pi\le 2x \le 4\pi,$$ $$-4\pi\le -2x \le -3\pi,$$ $$0\le 4\pi-2x \le \pi.$$

Let's break down the region into a few components.

1. When $3\le 4\pi-2x \le \pi$: The corresponding area is $\frac2{\pi}\cdot\frac{3(\pi-3)}2$
2. When $2 \le 4\pi-2x < 3$: The corresponding area is $\frac2{\pi}\cdot2\cdot \frac12$
3. When $1 \le 4\pi-2x < 2$: The corresponding area is $\frac2{\pi}\cdot1\cdot \frac12$.

The total area is $\frac2{\pi}\cdot \frac{3(\pi-3)+3}{2}=\frac{3(\pi-2)}{\pi}$

Verification using Wolfram Alpha: