Area of a mushroom-shaped curve

Inspired by a discussion on this question, I discovered the following hybrid function:

The curves in red are defined as $$f(x)=\exp\left((\sin x)^{(\sin x)^{\sin x}}\right)$$ and the curves in blue are defined as $$g(x)=(\sin x)^{\sin x}$$

The result looks like the head of a mushroom (with a bit of decoration :)

Question: Consider just one 'mushroom head'. What is the area?

We can rewrite the problem as $$\int_0^\pi\left[e^{(\sin x)^{(\sin x)^{\sin x}}}-(\sin x)^{\sin x}\right]\,dx$$ and we can see that it is symmetrical at $x=\pi/2$, since $\sin\left(\frac\pi2-x\right)=\sin\left(\frac\pi2+x\right)$, so this is equivalent to $$2\int_0^{\pi/2}\left[e^{(\sin x)^{(\sin x)^{\sin x}}}-(\sin x)^{\sin x}\right]\,dx\tag{1}$$

Wolfram Alpha calculates this definite integral to be around $3.88407$ (not equal to, as pointed out in the comments.

So how should I tackle this integral? I do not anticipate a closed form, hence approximations would be fine.

Update: I have approximated the functions into simpler ones, to give a value of $3.86029$.

Curve approximation

Over the interval $[0,\frac\pi2]$, the function $a(x)=\exp\left(\sin x^{\sin x^{\sin x}}\right)$ can be approximated by the function $$\alpha(x)=\frac53\sin x+1,$$ and similarly, the function $b(x)=\sin x^{\sin x}$ can be approximated by the function $$\beta(x)=\frac{0.85x\ln\left(0.66x\right)}{e^x}+1.$$ They are shown below, along with the original functions.

The function $\alpha$ is easy to integrate. We get $$\mathcal I_1=\int_0^{\pi/2}\alpha(x)\,dx=\left[x-\frac53\cos x\right]_0^{\pi/2}=\frac\pi2-\frac53.$$

The function $\beta$ is more difficult. Using WolframAlpha, we get $$\mathcal I_2=\int_0^{\pi/2}\beta(x)\,dx\approx1.30732.$$ To approximate this integral by hand, we could use the Taylor series for $\ln$ and $\exp$, but of course, this could only be limited to a number of terms (practically), as long rational functions are also extremely hard to integrate.

Hence the definite integral we want is $$\int_0^\pi a(x)-b(x)\,dx\approx2(\mathcal I_1-\mathcal I_2)=\pi+\frac{10}3-2\times1.30732=3.86029$$ with an error of around $0.61\%$.