Area bounded by $y=|\cos^{-1} (\sin x)|-|\sin^{-1} (\cos x)|$ The question is to evaluate the area bounded by $y=|\cos^{-1} (\sin x)|-|\sin^{-1} (\cos x)|$ and the $x$-axis for $x \in [3\pi/2,2\pi]$.
I tried to rewrite the equation as $y=|\cos^{-1} (\cos(\pi/2 - x))|-|\sin^{-1} (\sin (\pi/2-x))|$. Now I let $\pi/2-x=t$, then $-3\pi/2 \leq t \leq -\pi$. Accordingly $\cos^{-1} (\cos t)=t+2\pi$ and $\sin^{-1} (\sin t)=-t-\pi$. Putting this in the equation I got $y=4\pi-2x$. I am not getting the correct answer.
Another approach of mine was to rewrite is as $|\frac{\pi}{2} - \sin^{-1} \sin x|-|\frac{\pi}{2} - \cos^{-1}\cos x|$. And for $x\in [3\pi/2,2\pi]$, $\sin^{-1} \sin x=x-2\pi$ and $\cos^{-1} \cos x=-x+2\pi$. Putting this in the equation I got $y=4\pi-2x$, which is also giving incorrect result. Any ideas where I went wrong and what is the correct solution.

Your first approach should work, because indeed $y=4\pi-2x$. The integral is then $4\pi\frac{\pi}{2}-[x^2]^{2\pi}_{3\pi/2}=\frac{\pi^2}{4}$. (It's possible this is where you made a mistake.) When you say you're not getting the correct answer, what do you think that answer is? What's your source?
