Know Your Rules Backwards

$\begin{aligned} I & = \int_0^\frac \pi 4 e^{\sin x}(1+\tan x \sec x)\ dx & \small \color{#3D99F6} \text{Note that }\frac d{dx}e^{\sin x}\sec x = \cos xe^{\sin x}\sec x + e^{\sin x}\tan x \sec x \\ & = e^{\sin x}\sec x\ \bigg|_0^\frac \pi 4 & \small \color{#3D99F6} = e^{\sin x}(1+\tan x \sec x) \\ & = \sqrt 2 e^{\frac 1{\sqrt 2}} - 1 \\ & \approx \boxed{1.868} \end{aligned}$

Applying Integration by Part for the following integral, we have $\int^{\frac{\pi}{4}}_0\,e^{sin x}\,tan x\,sec x\,dx=[e^{sin x}\,sec x]^{\frac{\pi}{4}}_0-\int^{\frac{\pi}{4}}_0\,sec x\,e^{sin x}\,cos x\,dx=[e^{sin x}\,sec x]^{\frac{\pi}{4}}_0-\int^{\frac{\pi}{4}}_0\,e^{sin x}\,dx$

Move all integrals to the left side, we have $\int^{\frac{\pi}{4}}_0\,e^{sin x}(1+tan x\,sec x)\,dx=[e^{sin x}\,sec x]^{\frac{\pi}{4}}_0=e^{\frac{1}{\sqrt{2}}}\sqrt{2}-1=\boxed{1.868187713}$

$\begin{aligned} I &= \displaystyle \int_0^{\pi/4} e^{\sin x} (1 + \tan x \sec x)\, dx \\ \\ &= \displaystyle \int_0^{\pi/4} e^{\sin x}\, dx + \displaystyle \int_0^{\pi/4} e^{\sin x} \tan x \sec x\, dx \\ \\ &= \displaystyle \int_0^{\pi/4} e^{\sin x}\, dx + \sec x e^{\sin x} |_0^{\pi/4} - \displaystyle \int_0^{\pi/4} e^{\sin x}\, dx \\ \\ &= \sqrt{2}e^{\sqrt{2}/2} - 1 \\ \\ &\approx 1.868 \end{aligned}$