Let's do some calculus! (22)

$\begin{aligned} I & = \int_{-\pi/2}^{\pi/2} \frac {e^{|\sin x|}\cos x}{1+e^{\tan x}} dx \\ & = \int_{-\pi/2}^0 \frac {e^{\color{#D61F06}{-}\sin \color{#3D99F6}{x}}\cos \color{#3D99F6}{x}}{1+e^{\tan \color{#3D99F6}{x}}} d \color{#3D99F6}{x} + \int_0^{\pi/2} \frac {e^{\sin x}\cos x}{1+e^{\tan x}} dx & \small \color{#3D99F6}{\text{Let }u = -x, \ dx = - du} \\ & = \int^{\color{#3D99F6}{\pi/2}}_{\color{#3D99F6}{0}} \frac {e^{-\sin \color{#3D99F6}{(-u)}}\cos \color{#3D99F6}{(-u)}}{1+e^{\tan \color{#3D99F6}{(-u)}}} d \color{#3D99F6}{u} + \int_0^{\pi/2} \frac {e^{\sin x}\cos x}{1+e^{\tan x}} dx \\ & = \int^{\color{#3D99F6}{\pi/2}}_{\color{#3D99F6}{0}} \frac {e^{\sin \color{#3D99F6}{u}}\cos \color{#3D99F6}{u}}{1+e^{-\tan \color{#3D99F6}{u}}} d \color{#3D99F6}{u} + \int_0^{\pi/2} \frac {e^{\sin x}\cos x}{1+e^{\tan x}} dx & \small \color{#3D99F6}{\text{Putting }u=x } \\ & = \int_0^{\pi/2} e^{\sin x}\cos x \left(\frac 1{1+e^{-\tan x}} + \frac 1{1+e^{\tan x}} \right) dx \\ & = \int_0^{\pi/2} e^{\sin x}\cos x \left(\cancel{\frac {e^{\tan x}}{e^{\tan x}+1} + \frac 1{1+e^{\tan x}}}^1 \right) dx \\ & = \int_0^{\pi/2} e^{\sin x}\cos x \ dx \\ & = e^{\sin x}\bigg|_0^{\pi/2} = e - 1 \approx \boxed{1.718} \end{aligned}$