$$\frac{\pi}{4}-\frac{1}{2}\le \int_0^1{\frac{\arcsin x}{1+x^8}}\le \frac{\pi}{2}-1$$
I can say that I have close to zero experience with integral inequalities. I've searched for couple of theorems most of them involving monotonous function on $[0,1]$. Arcsin is increasing in this case and the other one is decreasing. You could say we have a product of increasing and decreasing function on this interval however I didn't find any theorems that cover this case. Also this integral doesn't seem to have elementary function.
Another thing I noticed is that most of the theorems cover only 1 part of this inequality; either $\le or \ge$ so my guess is we need some type of combination or perhaps find a new integral that is equal to $\frac{\pi}{4}-\frac{1}{2}$ and another that is equal to $\frac{\pi}{2}-1$.
What's the correct approach when we need to prove that the value of certain integral is in a provided interval?
For the bound on the right, $$ \int_0^1{\frac{\arcsin x}{1+x^8}}\,dx\leq \int_0^1\arcsin x\,dx=\frac\pi2-1. $$ For the bound on the left, $$ \int_0^1{\frac{\arcsin x}{1+x^8}}\,dx\geq\frac12\,\int_{0}^1{{\arcsin x}}\,dx=\frac12\,\left(\frac\pi2-1\right)=\frac\pi4-\frac12. $$