Are you sure about the sine integral?

First, note simply that, except for rather specific values of $A$ and $B$ , there is no way to express these integrals with elementary functions, but that doesn't mean one cannot solve the problem.

Initially, looking at $I_2$ , one might try substituting something like $x=\sin\left(u\right)$ (which will cancel out the $\sqrt{1-x^2}$ in the denominator), but ultimately it should be apparent that substituting $x=\sin\left(\frac{\pi u}{4}\right)$ in $I_2$ works best:

$\begin{aligned} I_2&=\frac{4^B}{{\pi}^B}\int_0^1 \frac{{\left(\arcsin\left(x\right)\right)}^B}{x^A\sqrt{1-x^2}}\ dx\\ &=\frac{\pi}{4}\int_0^2 \frac{u^B}{{\left(\sin\left(\frac{\pi u}{4}\right)\right)}^A\sqrt{1-\sin^2\left(\frac{\pi u}{4}\right)}}\cos\left(\frac{\pi u}{4}\right)\ du\\ &=\frac{\pi}{4}\int_0^2 \frac{u^B}{{\left(\sin\left(\frac{\pi u}{4}\right)\right)}^A}\ du\\ \end{aligned}$

Now, note the similarities between $I_1$ and $I_2$ . Then, if $\large f\left(x\right)=\frac{x^B}{{\left(\sin\left(\frac{\pi x}{4}\right)\right)}^A}$ ,

$\begin{aligned} I_1+I_2&=\frac{\pi}{4}\int_0^2 \frac{{\left(\sin\left(\frac{\pi x}{4}\right)\right)}^A}{x^B}\ dx+\frac{\pi}{4}\int_0^2 \frac{x^B}{{\left(\sin\left(\frac{\pi x}{4}\right)\right)}^A}\ dx\\ &=\frac{\pi}{4}\int_0^2 \left(f\left(x\right)+\frac{1}{f\left(x\right)}\right)\ dx\\ &\ge \frac{\pi}{4}\int_0^2 2\ dx\\ &= \pi \end{aligned}$

using the AM-GM inequality and the fact that $f\left(x\right)>0$ for $0< x< 2$ . Now, equality occurs when and only when $f\left(x\right)=\frac{1}{f\left(x\right)}$ for all $x$ . This is true when $A=B=0$ .