TITLE: $\int_{0}^{\pi} \sin{(x)}^{\cos{(x)}} dx$
QUESTION [5 upvotes]: The first problem I encountered $$\int_{0}^{\pi} \sin{(x)}^{\cos{(x)}} dx$$ It $\displaystyle\int_{-\infty}^{\infty} e^{-x^2} dx$ I tried to make it variable interchangeable, like in your problem, but I didn't get any results.
How can you prove that a function has no closed form integral?
Then I learned a lot through this post. But still
$$\int_{a}^{b} f(x)^{g(x)} dx$$
$$\begin{array}{I|l|l|I}f(x)&g(x)\\ \hline \sin{(x)}&\cos{(x)} \\ \cos{(x)}&\sin{(x)} \\ \tan{(x)}&\cot{(x)} \\ \arcsin{(x)}&\arccos{(x)} \\ \tan{(x)} &\sin{(x)} \\ ...&...\end{array}$$
I realized I didn't know how to calculate specific integrals of its type. I'm looking for a solution to your first problem, can you help me?
Can anyone help in $\int_0^\pi (\sin x )^{\cos x} dx$?
It's the same problem here, but it looks like there's been no answer. That's why I wanted to ask you again.
$WolframAlpha$ although he says there is a value to a particular integrale
$$\lim_{x\to\pi^{-}} \sin{(x)}^{\cos{(x)}}=\exp{(-\log{0})}=+\infty$$
is.

REPLY [3 votes]: As already said, you cannot even compute this integral over this range since
$$[\sin(x)]^{\cos(x)}=x-x^3 \left(\frac{1}{2}\log (x)+\frac{1}{6}\right)+\cdots$$ So, no problem around $x=0$. But
$$[\sin(x)]^{\cos(x)}=\frac 1{\pi-x}+(\pi-x)\left(\frac{1}{2}\log (\pi-x)+\frac{1}{6}\right)+\cdots$$ and then a very serious problem close the upper bound which is the pole.
Now, using numerical integration for
$$f(k)=\int_0^{\pi-10^{-k}} [\sin(x)]^{\cos(x)}\,dx$$ you would have
$$\left(
\begin{array}{cc}
 k & f(k) \\
 0 & 1.64363 \\
 1 & 3.89600 \\
 2 & 6.19254 \\
 3 & 8.49501 \\
 4 & 10.7976 \\
 5 & 13.1002 \\
 6 & 15.4028
\end{array}
\right)$$