Integrate with series

$\begin{aligned} f(x) & = \sum_{n=0}^\infty \frac{(-x^2)^n}{(2n+1)!} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{x(2n+1)!} = \frac{\sin{x}}{x} \end{aligned}$

$\begin{aligned} \int_0^\frac{\pi}{2} f(x) f\left(\frac{\pi}{2} - x \right) dx & = \int_0^\frac{\pi}{2} \frac{\sin{x}}{x}\dot{} \frac{\sin{\left(\frac{\pi}{2} - x\right)}}{\frac{\pi}{2} - x} dx \\ & = \int_0^\frac{\pi}{2} \frac{\sin{x}\cos{x}}{x\left(\frac{\pi}{2} - x\right)} dx \\ & = \int_0^\frac{\pi}{2} \frac{\sin{(2x)}}{x\left(\pi - 2x \right)} dx \quad \quad \small \color{#3D99F6}{\text{Let } u = 2x \quad \Rightarrow dx = \frac{1}{2} du} \\ & = \int_0^\pi \frac{\sin{(u)}}{u\left(\pi - u \right)} du \quad \quad \small \color{#3D99F6}{\text{We note that: } \frac{1}{u} + \frac{1}{\pi-u} = \frac{\pi}{u(\pi-u)}} \\ & = \frac{1}{\pi} \int_0^\pi \left(\frac{1}{u}+\frac{1}{\pi-u}\right)\sin{u} \space du \quad \quad \small \color{#3D99F6}{\text{Since: } \sin{(\pi-u)} = \sin{u}} \\ & = \frac{1}{\pi} \int_0^\pi \left(\frac{\sin{u}}{u} + \frac{\sin{(\pi-u)}}{\pi - u}\right) du \quad \quad \small \color{#3D99F6}{\text{Let } v = \pi - u \quad \Rightarrow du = -dv} \\ & = \frac{1}{\pi} \left( \int_0^\pi \frac{\sin{u}}{u} du - \int_\pi^0 \frac{\sin{v}}{v} dv \right) \\ & = \frac{1}{\pi} \left( \int_0^\pi \frac{\sin{u}}{u} du + \int_0^\pi \frac{\sin{v}}{v} dv \right) \\ & = \frac{2}{\pi} \int_0^\pi f(x) dx \end{aligned}$

$\Rightarrow A + B = 2 + 1 = \boxed{3}$