You should look at the graphs of these functions

Consider the integral $\displaystyle I = \int_0^\infty e^{x^2i} \cos x \ dx = \int_0^\infty \cos x \cos x^2 \ dx + i \int_0^\infty \cos x \sin x^2 \ dx$ . Implying that the real part $\displaystyle \Re(I) = \int_0^\infty \cos x \cos x^2 \ dx$ and the imaginary part $\displaystyle \Im(I) = \int_0^\infty \cos x \sin x^2 dx$ .

$\begin{aligned} I & = \int_0^\infty e^{x^2i} \cos x \ dx = \int_0^\infty e^{x^2i} \cdot \frac {e^{xi} + e^{-xi}}2 dx \\ & = \frac 12\left(\int_0^\infty e^{(x^2+x)i} dx + \int_0^\infty e^{(x^2-x)i} dx\right) \\ & = \frac 12\left(\int_0^\infty e^{\left(\left(x+\frac 12\right)^2 - \frac 14\right)i} dx + \int_0^\infty e^{\left(\left(x-\frac 12\right)^2 - \frac 14\right)i} dx\right) & \small \color{#3D99F6} \text{Let }u = x+\frac 12, v = x - \frac 12 \\ & = \frac {e^{-\frac i4}}2\left(\int_{\frac 12}^\infty e^{u^2i} du + \int_{-\frac 12}^\infty e^{v^2i} dv \right) \\ & = \frac {e^{-\frac i4}}2\left(\int_0^\infty e^{u^2i} du + \int^{\frac 12}_0 e^{u^2i} du + \int_{-\frac 12}^0 e^{v^2i} dv + \int_0^\infty e^{v^2i} dv \right) \\ & = \frac {e^{-\frac i4}}2\left(\int_0^\infty e^{u^2i} du + \int^{\frac 12}_0 e^{u^2i} du - \int_0^{\frac 12} e^{v^2i} dv + \int_0^\infty e^{v^2i} dv \right) \\ & = e^{-\frac i4} \int_0^\infty e^{\color{#3D99F6}u^2i} du & \small \color{#3D99F6} \text{Let }-t^2 = u^2 i \implies e^{\frac \pi 4 i} dt = du \\ & = e^{\frac {\pi - 1}4i} {\color{#3D99F6} \int_0^\infty e^{-t^2} dt} = e^{\frac {\pi - 1}4i} \cdot {\color{#3D99F6}\frac {\sqrt \pi}2\text{ erf }(\infty)} = \frac {\sqrt \pi e^{\frac {\pi - 1}4i}}2 & \small \color{#3D99F6} \text{Error function erf }(\infty) = 1 \\ & = \frac {\sqrt \pi}2 \left(\cos \frac \pi 4 + i\sin \frac \pi 4\right) \left(\cos \frac 14 - i\sin \frac 14\right) \\ & = \sqrt {\frac \pi 8} \left(1 + i\right) \left(\cos \frac 14 - i\sin \frac 14\right) \\ & = \sqrt {\frac \pi 8} \left(\cos \frac 14 + \sin \frac 14 + i\left(\cos \frac 14 - \sin \frac 14 \right) \right) \end{aligned}$

Then we have $\displaystyle \int_0^\infty \cos x \cos x^2 \ dx \int_0^\infty \cos x \sin x^2 \ dx = \Re(I) \Im(I) = \frac \pi 8 \left( \cos^2 \frac 14 - \sin^2 \frac 14 \right) = \frac \pi 8 \cos \frac 12$ . Therefore, $A+B = 8 + 2 = \boxed{10}$ .