Integral Bee Preparation -- Trouble with this beautiful integral

[YAYA12345]

TL;DR Summary

While I was preparing for an integrals contest, I had a doubt about the following integral, I tried several substitutions but nothing worked.I would appreciate your support for this beautiful integral.
$$ \int\limits_{0}^{1/2} \cos(1-\cos(1-\cos(...(1-\cos(x))...) \ \mathrm{d}x$$

While I was preparing for an integrals contest, I had a doubt about the following integral, I tried several substitutions but nothing worked.I would appreciate your support for this beautiful integral.
$$ \int\limits_{0}^{1/2} \cos(1-\cos(1-\cos(...(1-\cos(x))...) \ \mathrm{d}x$$

 

[fresh_42]

TL;DR Summary: While I was preparing for an integrals contest, I had a doubt about the following integral, I tried several substitutions but nothing worked.I would appreciate your support for this beautiful integral.
$$ \int\limits_{0}^{1/2} \cos(1-\cos(1-\cos(...(1-\cos(x))...) \ \mathrm{d}x$$

While I was preparing for an integrals contest, I had a doubt about the following integral, I tried several substitutions but nothing worked.I would appreciate your support for this beautiful integral.
$$ \int\limits_{0}^{1/2} \cos(1-\cos(1-\cos(...(1-\cos(x))...) \ \mathrm{d}x$$

Looks as if your integrand converges to identical $1.$

 

[YAYA12345]

¿Cómo lo sabes?
Translation by mentor: How do you know that?

@YAYA12345 -- per the forum rules, posts must be in English.

 

[fresh_42]

¿Cómo lo sabes?

https://www.wolframalpha.com/input?i=integral+(from+0+to+(1/2))++cos(1-cos(1-cos(x)))+dx+=

and

https://www.wolframalpha.com/input?i=integral+(from+0+to+(1/2))++cos(1-cos(1-cos(1-cos(1-cos(x)))))+dx+=

I wanted to know what had to be proven.

 

[YAYA12345]

https://www.wolframalpha.com/input?i=integral+(de+0+a+(1/2))++cos(1-cos(1-cos(x)))+dx+=

y

https://www.wolframalpha.com/input?i=integral+(de+0+a+(1/2))++cos(1-cos(1-cos(1-cos(1-cos(x))) )))+dx+=

Quería saber qué había que probar. :Frío:

¿Y cómo podría probarlo? ¿Puedes apoyarme con una pista?

(Approximate) Translation by mentor -- And how could it be proven? Can you lead me to a track?

 

[fresh_42]

¿Y cómo podría probarlo? ¿Puedes apoyarme con una pista?

I suggest we continue in English.

I would start with the Taylor expansion:
\begin{align*}
\cos(x_0) &= 1- \dfrac{x_0^2}{2!}+\dfrac{x_0^4}{4!}-\dfrac{x_0^6}{6!}\pm \ldots \\
1-\cos(x_0) &= \dfrac{x_0^2}{2!}-\dfrac{x_0^4}{4!}+\dfrac{x_0^6}{6!}\mp \ldots =:x_1\\
\cos(x_1) &= 1 - \dfrac{x_1^2}{2!}+\dfrac{x_1^4}{4!}-\dfrac{x_1^6}{6!}\pm \ldots \\
&=1- O(x_0^4) \\
& etc.
\end{align*}
The second term is already small on $I:=[0\, , \,0.5].$ Iteration looks as if
$$
cos(1-cos(1-cos(1-cos(1-cos(x))))) = 1 + O\left(\left(\dfrac{x}{2}\right)^{2^5}\right)
$$
or similar. The big-O term quickly becomes negligible on $I.$

 

[Mark44]

My approach would be to calculate a few integrals of increasing complexity; i.e., $\int \cos(1 - \cos(x))dx, \int \cos(1 - \ cos(1 - \cos(x))))dx$, and maybe one more to see if a pattern emerges. If so, I would then try a proof by induction. I can't guarantee this would work, but it's the direction I would start with.

 

[fresh_42]

¿Y cómo podría probarlo? ¿Puedes apoyarme con una pista?

(Approximate) Translation by mentor -- And how could it be proven? Can you lead me to a track?

Another idea:

Let's define $f_0(x)=\cos(x)$ and $f_n(x)=\cos(1-f_{n-1}(x)).$ We want to show that $L(x):=\lim_{n \to \infty}f_n(x) \equiv 1.$

We have
\begin{align*}
L(x)&=\lim_{n \to \infty}f_n(x)\\&=\lim_{n \to \infty}(\cos(1-f_{n-1}(x)))\\
&=\cos(1-\lim_{n \to \infty}f_{n-1}(x))\\&=\cos(1-L(x))
\end{align*}
and so $L(x)=\cos(1-L(x)).$

So you only have to show that $x=\cos(1-x)$ has only the solution $x=1.$

 

[Office_Shredder]

So you only have to show that $x=\cos(1-x)$ has only the solution $x=1.$

As long as we're doing calculus: x and $\cos(1-x)$ are both increasing on $[0,1]$ (it can't be a negative x because cosine is always positive on $[-1,1]$) and the slope of cosine is strictly less than 1 on the interval, so they can only intersect at most once.

I don't know what any of this has to do with an integration bee though.

 

[fresh_42]

As long as we're doing calculus: x and $\cos(1-x)$ are both increasing on $[0,1]$ (it can't be a negative x because cosine is always positive on $[-1,1]$) and the slope of cosine is strictly less than 1 on the interval, so they can only intersect at most once.

I don't know what any of this has to do with an integration bee though.

I tried to leave at least a bit for the OP to do.

 

[Infrared]

So you only have to show that $x=\cos(1-x)$ has only the solution $x=1.$

This is the approach that I would follow too, but there is one more needed step: to explain why the limit $L(x)$ exists in the first place. I think the easiest argument would be to notice that the map $t\mapsto cos(1-t)$ is a contraction and apply contraction-mapping. Usually you need the contraction to decrease the distance by a factor bounded away from 1 but since you can take the domain to be compact it's enough that it strictly decreases distance between distinct points.