# Show that $\int_0^1 {\sin \left( {\pi x} \right)} x^x \left( {1 - x} \right)^{1 - x} \mathrm dx = \frac{{\pi e}}{{24}}$ [duplicate]

While going through a list (Problem $$17$$) of interesting integrals I have come across this one

$$\int_0^1 {\sin \left( {\pi x} \right)} x^x \left( {1 - x} \right)^{1 - x} \mathrm dx = \frac{{\pi e}}{{24}}$$

I have tried integration by parts with $$u=x^x \left( {1 - x} \right)^{1 - x}$$ and $$\mathrm dv=\sin \left( {\pi x} \right)$$ but this ended up in and even more complicated integral. To use the series representation of $$e^{x\ln(x)}$$ and respectively $$e^{(1-x)\ln(1-x)}$$ does not appeared to be helpful at all even after reshaping it all in the form $$(1-x)\left(\frac{x}{1-x}\right)^x$$. Then I thought about using Euler's Reflection Formula to get rid of the $$\sin(\pi x)$$ and work in terms of the Gamma Function instead. Hence this formula only holds for $$x\notin\mathbb{Z}$$ $$-$$ and the limits are given by $$0$$ and $$1$$ $$-$$ I guess this is not possible here. My last try was to use the Weierstrass Expansion of the sine function but I guess this is not the right approach either.

Could someone please provide a whole solution since I have no further idea how to deal with this integral? I have searched here on MSE but it does not seem like someone hast asked something like this before. Nevertheless tell me when I have overseen something or when you can link a former question which is helpful for understanding the process of evaluating this definite integral.