A calculus problem by Akeel Howell

We can rewrite $f(x)$ by solving the integral. If we let $u = \arcsin(x), du = \dfrac{1}{\sqrt{1-x^2}}dx$ and the integral becomes $\displaystyle{f(x) = \pi\int{e^{2u}\cos{u}}du}$ . We can then proceed by integration by parts to get

$\displaystyle{\pi\int{e^{2u}\cos{u}}du} = \pi \left(e^{2u}\sin{u}-2\int{\sin{u}e^{2u}}du\right)+C \\ = \pi \left(e^{2u}\sin{u}-2\left[-e^{2u}\cos{u}+\int{2e^{2u}\cos{u}du}\right]\right)+C$ $= \pi \left(e^{2u}\sin{u}+2e^{2u}\cos{u}-4\int{e^{2u}\cos{u}du}\right)+C \\ \implies \int{e^{2u}\cos{u}du} = \dfrac{1}{5}\left(e^{2u}\sin{u}+2e^{2u}\cos{u}\right)+C$ $\cos{(\arcsin{x})} = \sqrt{1-x^2} \implies \dfrac{1}{5}\left(e^{2u}\sin{u}+2e^{2u}\cos{u}\right) = \dfrac{1}{5}\left(e^{2\arcsin{x}}x+2e^{2\arcsin{x}}\sqrt{1-x^2}\right)+C$ $\therefore f(x) = \dfrac{\pi}{5}e^{2\arcsin{x}}\left(x+2\sqrt{1-x^2}\right)+C \\ \text{Since } \space f(0) = \dfrac{2\pi}{5} \implies C = 0$ $\text{Let } \space \arcsin(3) = y \implies \sin(y) = 3 \\ \text{So } \space e^{iy}-e^{-iy} = 6i$ $\therefore e^{iy} = \dfrac{6i \pm \sqrt{-32}}{2} = 3i \pm 2i\sqrt{2}$ $\text{So } \space iy = \ln{i(3 \pm 2\sqrt{2})} \implies y = \dfrac{\pi}{2}-i\ln{(3 \pm 2\sqrt{2})}$ $\text{So } \space f(3) = \dfrac{\pi}{5}e^{\pi-2i\ln{(3 \pm 2\sqrt{2})}}(3+2\sqrt{-8})$ $\therefore |f(3)| = \left|\dfrac{\pi}{5}e^{\pi-2i\ln{(3 \pm 2\sqrt{2})}}(3+2\sqrt{-8})\right| \\ = \dfrac{\pi}{5}e^{\pi}\left|(3+2\sqrt{-8})\right| = \dfrac{\sqrt{41}}{5}\pi e^{\pi}$ $\therefore a+b = 41+5 = \space \boxed{46}$