By parts

$\begin{aligned} f \left(\frac {\color{#3D99F6}\arcsin x}2\right) & = 4e^{\color{#3D99F6}\arcsin x} x + \arccos \left(\frac \pi 2 - {\color{#D61F06}\arccos x}\right) + 3 & \small \color{#3D99F6} \text {Let } \arcsin x = \theta \implies \sin \theta = x \\ \implies f\left(\frac \theta 2\right) & = 4e^{\color{#3D99F6}\theta} \sin \theta + \arccos \left(\frac \pi 2 - {\color{#D61F06}\left(\frac \pi 2 - \theta\right)}\right) + 3 & \small \color{#D61F06} \text {Note that } \arccos x = \frac \pi 2 - \arcsin x \\ & = 4e^\theta \sin \theta + {\color{#3D99F6}\arccos \theta} + 3 & \small \color{#3D99F6} \text {Let } \arccos \theta = \phi \implies \cos \phi = \theta \\ \implies f \left(\frac {\cos \phi}2\right) & = 4e^{\cos \phi} \sin (\cos \phi) + \phi + 3 & \small \color{#3D99F6} \text {Let } 2u = \cos \phi \\ \implies f(u) & = 4e^{2u} \sin (2u) + \arccos (2u) + 3 \end{aligned}$

Then we have:

$\begin{aligned} \int_{-\frac 12}^\frac 12 f(x) \ dx & = \int_{-\frac 12}^\frac 12 4e^{2x} \sin (2x) \ dx +\int_{-\frac 12}^\frac 12 \arccos (2x)\ dx +\int_{-\frac 12}^\frac 12 3 \ dx & \small \color{#3D99F6} \text {Let } t = 2x \\ & = {\color{#3D99F6}2 \int_{-1}^1 e^t \sin t \ dt + \frac 12 \int_{-1}^1 \arccos t\ dt} + 3x \ \bigg|_{-\frac 12}^\frac 12 & \small \color{#3D99F6} \text {By integration by parts} \\ & = {\color{#3D99F6}2 \left(-e^t \cos t + \int e^t \cos t \ dt \right) + \frac 12 \left(t\arccos t + \int \frac t{\sqrt{1-t^2}}dt \right) \bigg|_{-1}^1} + 3 \\ & = {\color{#3D99F6}2 \left(-e^t \cos t + e^t \sin t - \int e^t \sin t \ dt \right) + \frac 12 \left(t\arccos t - \sqrt{1-t^2} \right) \bigg|_{-1}^1} + 3 \\ & = e^t (\sin t - \cos t)\ \bigg|_{-1}^1 + \frac \pi 2 + 3 \\ & = e(\sin 1 - \cos 1)+e^{-1}(\sin 1+\cos 1) + \frac \pi 2 + 3 \\ & \approx \boxed{5.898} \end{aligned}$