$\int_{0}^{\pi} \sin{(x)}^{\cos{(x)}} dx$

Question

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.

If you have trouble writing in English, try using Google Translate before posting. — jjagmath, Sep 22 '21 at 14:49
@ThomasAndrews I wanted to adapt a similar expression to it because we could calculate a specific integrali, even though it didn't have an ambiguous integrali. — Bell Cranel, Sep 22 '21 at 15:33
Actually the second post that you linked to does answer your question, since there is an answer there showing that the integral is divergent. What more could you ask for? — Hans Lundmark, Sep 22 '21 at 17:35
On the other hand:$$\int_0^\pi \sin(x)^{\sin(x)}dx$$ and $$\int_0^\frac{\pi}{2}\sin(x)^{\cos(x)} dx $$ and others converge. — Тyma Gaidash, Sep 23 '21 at 12:49

score 3 · Accepted Answer · answered Sep 23 '21 at 03:28

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 $$\left( \begin{array}{cc} k & f(k) \\ 0 & 1.64363 \\ 1 & 3.89600 \\ 2 & 6.19254 \\ 3 & 8.49501 \\ 4 & 10.7976 \\ 5 & 13.1002 \\ 6 & 15.4028 \end{array} \right)$$

$\int_{0}^{\pi} \sin{(x)}^{\cos{(x)}} dx$

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