Although not this particular problem, I've looked about evaluating an analogous integral in hopes of furthering my understanding what real methods would be of good use.

Consider the following integral:

$$ \int_0^1 x^x (1-x)^{1-x} \sin(\pi x) \mathrm{d} x$$

Using the method of residues, the integral can be shown to equal $\frac{\pi e}{24}$. However, here's one real attempt. Consider the following Gamma function identity

$$\Gamma(x) = \sqrt{2 \pi } x^{x-1/2} e^{-x} e^{f(x)} $$

where $$f(x) = \int_0^\infty \frac{2\arctan{\frac{t}{x}}}{e^{2\pi t}-1} \mathrm{d}t$$

Using Euler's reflection formula $\Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin{\pi x}}$, we have

$$ \int_0^1 x^x (1-x)^{1-x} \sin(\pi x) \mathrm{d} x = \frac{e}{2} \int_0^1 \sqrt{x(1-x)}e^{-(f(x)+f(1-x))} \mathrm{d}x$$

Substituting $x = \sin^2(\frac{\pi\theta}{2})$, we have then

$$\frac{e}{2} \int_0^1 \sqrt{x(1-x)}e^{-(f(x)+f(1-x))} \mathrm{d}x = \frac{\pi e}{8}\int_0^1 \sin^2(\pi \theta) e^{-H(\theta)}\mathrm{d} \theta$$

where $$ H(\theta) = f\left(\sin^2\left(\frac{\pi\theta}{2}\right)\right) + f\left(\cos^2\left(\frac{\pi\theta}{2}\right)\right) = \int_0^\infty \frac{2\arctan{\frac{4t}{\sin^2(\pi \theta)-4t^2}}}{e^{2\pi t}-1} \mathrm{d}t$$

I'm still looking towards it, so at least we can claim the original integral is bounded by $\frac{\pi e}{16} $ by using the fact $e^{-H(\theta)}$ is positive, and $\int_0^1 \sin^2(\pi \theta) \mathrm{d} \theta = \frac{1}{2}$. Thus, our integral reduces to showing the following

$$\int_0^1 \sin^2(\pi \theta) e^{-H(\theta)}\mathrm{d} \theta = \frac{1}{3}$$

using real methods.

I'll continue updating as I find ways to approach this problem.