Oh no!

So this is one of misconceptions on trigonometry. The range of $\arcsin$ is $[-\pi/2,\pi/2]$ , and the range of $\arccos$ is $[0,\pi]$ . For example, $\arcsin(\sin x)=x$ if $x \in [-\pi/2,\pi/2]$ . Otherwise, we need to derive something. For $-\pi/2 \le x \le \pi$ , we know that $\sin x=\sin(\pi-x)$ , where $\pi-x \in [-\pi/2,\pi/2]$ . It means that for this case, $\arcsin(\sin x)=\pi-x$ . Similarly, for $-\pi \le x \le -\pi/2$ , we have $\arcsin(\sin x)=-\pi-x$ . With the same trick, we have $\arccos(\cos x)=x$ if $x \in [0,\pi]$ , and $\arccos(\cos x)=-x$ if $x\in[-\pi,0]$ .

The hard parts are two other functions, $\arccos(\sin x)$ and $\arcsin(\cos x)$ . For the first one, if $-\pi \le x \le -\pi/2$ , we have $\sin x=\cos(\pi/2-x)=\cos(x-\pi/2)=\cos(x+3\pi/2)$ , where $x+3\pi/2 \in [\pi/2,\pi]$ . So we have that $\arccos(\sin x)=x+3\pi/2$ . I won't do for every interval, because it is kind of messy stuff. Here's a conclusion:

$\begin{matrix} {\rm Function} & [-\pi,-\pi/2] & [-\pi/2,0] & [0,\pi/2] & [\pi/2,\pi]\\ \arcsin(\sin x) & -x-\pi & x & x & -x+\pi\\ \arccos(\cos c) & -x & -x & x & x\\ \arcsin(\cos x) & x+\pi/2 & x+\pi/2 & -x+\pi/2 & -x+\pi/2\\ \arccos(\sin x) & x+3\pi/2 & -x+\pi/2 & -x+\pi/2 & x-\pi/2 \end{matrix}$

If we denote the integrand function by $F(x)$ , we get

$F(x)=\begin{cases} x^4+3\pi x^3+\frac{11\pi^2}{4}x^2+\frac{3\pi^3}{4}x & -\pi \le x < -\frac{\pi}{2},\\ x^4-\frac{\pi^2}{4}x^2 & -\frac{\pi}{2} \le x < 0,\\ x^4-\pi x^3+\frac{\pi^2}{4}x^2 & 0 \le x < \frac{\pi}{2},\\ x^4-2\pi x^3+\frac{5\pi^2}{4}x^2-\frac{\pi^3}{4}x & \frac{\pi}{2} \le x \le \pi. \end{cases}$

Integrate each parts and combine them, then the integral equals $\frac{\pi^5}{240}$ . Thus the answer is $\frac{240}{5}=48$ .