A Very Odd Integral

Let $f(x)=x^3 \sec^2 x$ . Then $f(-x)={(-x)}^{3} \sec^2 {(-x)} = -(x^3 \sec^2 x) = -f(x)$ . Therefore the function is odd and since the function is defined for $x \in [ -\frac{\pi}{4}, \, \frac{\pi}{4} ]$ the integral is equal to $0$ .

We need to find two integers $A$ and $B$ such that ${\pi}^{A}-{B}=0$ , and the only way for this to be possible is if $A=0$ (any other integer will include pi), and therefore $B=1$ .

Therefore $A+B=0+1=\large \color{#20A900}{\boxed{1}}$ .