Integration of a Exponential Function with a Trigonometric argument

We cannot expect an explicit antiderivative but $$x^{\sin(x)}=e^{\sin(x)\log(x)}$$ Now, compose Taylor series around $x=0$ to have $$x^{\sin(x)}=\sum_{n=0}^\infty \frac{a_n}{n!} x^n$$ where the $a_n$ are polynomials in $t=\log(x)$.

Computing the first ones $$\left( \begin{array}{cc} n & a_n \\ 0 & 1 \\ 1 & t \\ 2 & t^2 \\ 3 & t^3-t \\ 4 & t^4-4 t^2 \\ 5 & t^5-10 t^3+t \\ 6 & t^6-20 t^4+16 t^2 \\ 7 & t^7-35 t^5+91 t^3-t \\ 8 & t^8-56 t^6+336 t^4-64 t^2 \\ 9 & t^9-84 t^7+966 t^5-820 t^3+t \\ 10 & t^{10}-120 t^8+2352 t^6-5440 t^4+256 t^2 \end{array} \right)$$ which means that for $n \ge 1$, you face integrals $$I_{m,n}=\int_0^1 x^m \,\log ^n(x)\,dx=(-1)^n \frac {n!}{(m+1)^{n+1} }$$

Using the above table, as a function of the order of the expansion we should have the successive approximations for $$\int_0^1 x^{\sin(x)}\,dx$$ $$\left( \begin{array}{ccc} 0 & 1 & 1.000000 \\ 1 & \frac{3}{4} & 0.750000 \\ 2 & \frac{85}{108} & 0.787037 \\ 3 & \frac{5485}{6912} & 0.793547 \\ 4 & \frac{17089937}{21600000} & 0.791201 \\ 5 & \frac{461505799}{583200000} & 0.791334 \\ 6 & \frac{380113636145857}{480290277600000} & 0.791425 \end{array} \right)$$ while numerical integration gives $0.791403$.

For the fun of it, I used all the values given in the first table and obtained $$\frac{11803390040762292561031266684259753655627}{14914487726878692033020558868480000000000}=\color{red}{0.7914043}22$$ to be compared to $\color{red}{0.791404300}$

For the case of $$\int_0^{\frac \pi 2} x^{\sin(x)}\,dx$$ the same could apply but it is more complicated since $$I_{m,n}=\int_0^{\frac \pi 2} x^m \,\log ^n(x)\,dx=(-1)^n \frac{\Gamma \left(n+1,(m+1) \log \left(\frac{2}{\pi }\right)\right) } {(m+1)^{n+1} }$$ where appears the incomplete gamma function.

Using all given terms, we should have $1.52012$ while numerical integration gives $1.52029$

Edit (for your curiosity)

It would have been nice to work with $$x^{\sin(x)}=\sum_{n=0}^\infty \frac{1}{n!} \big[ \sin(x)\log(x) \big]^n$$

$$\int x^{\sin(x)}\,dx=\sum_{n=0}^\infty \frac{1}{n!}\int \big[ \sin(x)\log(x) \big]^n\,dx$$ The integrals $$I_n=\int \big[ \sin(x)\log(x) \big]^n\,dx$$ are known but they involve a bunch of hypergeometric functions (even with imaginary coefficients).

The exception is the first one which is easy to compute using one integration by parts $$I=\int \sin(x)\log(x) \,dx=\text{Ci}(x)-\log (x) \cos (x)$$ from which $$I=\int_0^1 \sin(x)\log(x) \,dx=\text{Ci}(1)-\gamma$$ would give for the definite integral $0.760188$. $$I=\int_0^{\frac \pi 2} \sin(x)\log(x) \,dx=\text{Ci}\left(\frac{\pi }{2}\right)-\gamma$$ would give for the definite integral $1.46558$.