$\int_0^{1398\pi} ({\sin^2x})^{\cos^2x}dx=?$

Question

$$\int_0^{1398\pi} ({\sin^2x})^{\cos^2x}dx=?$$ $f(x)=({\sin^2x})^{\cos^2x}$ is periodic with period equal to $\pi$,therefore we need to compute the integral in one period. The original question was: $$\int_0^{1398\pi} ({\sin^2x})^{\cos^2x}+({\cos^2x})^{\sin^2x}dx=?$$Proposed by Jalil Hajimir to Romanian Math Magazine

Answer 1

Comment.

It's a bit more pleasant to use e.g. $\displaystyle ~\int\limits_0^1 \frac{x^{1-x}+(1-x)^x}{\sqrt{x(1-x)}}dx~$ than $\displaystyle ~\int\limits_0^{\pi} ((\sin^2 x)^{\cos^2 x} + (\cos^2 x)^{\sin^2 x}) dx~$ .

I don't know a closed form for that, and it's also very unlikely that exists one here.

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Note:

If we define $~p_n(x)~$ as a polynomial of degree $n$ , we get:

$$\int\limits_0^1 \frac{x^{1-x}+(1-x)^x}{\sqrt{x(1-x)}}dx = 2\sqrt{2} \sum\limits_{k=0}^{\infty}\frac{p_{2k}(\ln 2)}{2^{2k}(2k+1)!}$$

$p_0(x)=1$

$p_2(x)=x^2+2x-1$

$p_4(x)=x^4+4x^3-6x^2+60x+65$

$p_6(x)=x^6+6x^5-15x^4+300x^3+975x^2+3750x+6687$

$p_8(x)=$

$=x^8+8x^7-28x^6+840x^5+4550x^4+35000x^3+187236x^2+769272x+1305089$

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