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This is only about, how I came up with this weird idea.

I was considering about relationship between radius($r$),arc of the circle ($s$) and radian ($\theta$) such that;

enter image description here

$$\boxed{r/s=\theta}$$

And to convince myself completely, I wanted to try following method. When $s,\phi$ are very small like $\delta s$ and $\delta \phi$, we can assume that the triangle that $rsr$ with angle $\delta \phi$ that is; enter image description here

So from the triangle, we can conclude that $$\sin\left(\frac{\delta \phi}{2}\right)=\dfrac{\delta s}{2r}$$ and while $\delta s,\delta \phi$ go to the $0$ we can take integral at both side.

Conclusion:

What does following mean?

$$\boxed{I=\displaystyle\int \sin(dx)}$$

1.

I considered that what if we take the integral to inside of $\sin(x)$?

so $I=\sin (x+C)$

2.

I've tried definition of riemann integral.

$$\displaystyle\int_a^b f(x)dx=\lim\limits_{n\to \infty}\displaystyle\sum_{k=0}^n f\left(a+k\dfrac{b-a}{n}\right)\left(\dfrac{b-a}{n}\right)$$

But what is the function? $f(dx)$ doesn't look like just $f(x)$ or I can try following but it doesn't make any sense to me, as well.

$$\displaystyle\int f(dx)=\int \dfrac{f(dx)}{dx}dx$$ so $U(x)=\dfrac{f(dx)}{dx}$, but I couldn't finish.

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    $\begingroup$ In terms of Riemann sums, I imagine you'd have: $$\int_a^bf(dx)=\lim_{n\to\infty}\sum_{k=0}^nf\left(\frac{b-a}n\right)=\lim_{n\to\infty}(n+1)f\left(\frac{b-a}n\right)$$Though this is just a guess. $\endgroup$ – Simply Beautiful Art Jul 15 '17 at 12:09
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    $\begingroup$ That's mostly meaningless $\endgroup$ – Hagen von Eitzen Jul 15 '17 at 12:10
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    $\begingroup$ just an idea: probably (not sure) $\mathrm dx$ is defined in non-standard analysis in terms of infinitesimals. If this is the case then we can find a meaning for some expression of the kind $f(\mathrm dx)$. $\endgroup$ – Masacroso Jul 15 '17 at 12:14
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    $\begingroup$ @HagenvonEitzen. Many ideas have been meaningless in the beginning, but later got a rigorous mathematical treatment, e.g. the Dirac $\delta$ "function". The idea in the question is not that stupid and actually can be given a meaning. $\endgroup$ – md2perpe Jul 15 '17 at 22:14
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    $\begingroup$ The most natural value of $f(dx)$ is $f'(0) \, dx$. Generally, $f(x+dx) = f(x) + f'(x) \, dx$. Therefore, $U(x) = f'(0)$. $\endgroup$ – md2perpe Jul 15 '17 at 22:17
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Perhaps this will help: for very small angles $\sin(x) = x$ is a very good approximation. You can use this to show: $\dfrac{\delta \phi}{2} =\dfrac{\delta s}{2r}$ which leads to $\delta \phi = \dfrac{\delta s}{r}$ and now you can integrate both sides to show $s=2\pi r$.

Note: I respectfully point out that arc-length is proportional to both radius and angle so: $ s = r \phi$; I think you might have a typo in your first boxed equation.

But I don't think I answered the question... what could $\int \sin(dx)$ mean.

One possible way to look at what $\sin(dx)$ means could come from the definition of the derivative: $f'(x)=\lim_{dx \to 0} \left[ \frac {f(x+dx)-f(x)} {dx} \right]$

now take some liberty with the limit and move it to the whole equation level. Also, evaluate the equation at zero.

$\lim_{dx \to 0} \left[ f'(0)= \frac {f(0+dx)-f(0)} {dx} \right]$

And rearrange to isolate $f(dx)$.

$\lim_{dx \to 0} \left[ f(dx)=f(0)+f'(0) dx \right]$

So: $\lim_{dx \to 0} \left[ \sin(dx) = \sin(0)+\cos(0)dx = dx \right]$

Which gets us back to the approximation that $\sin(x) = x$ for very small angles.

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With particular reference to the sector of circle given $$ \boxed{r = s \cdot \theta} $$ is incorrect, chasing it could lead to weird results. $$ \boxed{s = r\cdot \theta} $$ is correct, leading to correct result.

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The text that you reproduced does not mention the "conclusion"

$$\int\sin(dx)$$ which is meaningless (there isn't even a variable $x$ in the explanation).

What you can write is

$$\sin\left(\frac{\delta \phi}{2}\right)\approx\frac{\delta \phi}{2}=\dfrac{\delta s}{2r}$$

and from this, replacing small increments by differentials,

$$S=\oint ds=\int_0^\theta r\,d\phi=r\theta.$$

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