What does $\int \sin(dx)$ mean? 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;

$$\boxed{r/s=\theta}$$
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;
  
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.
 A: 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.
A: 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.
A: 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,
$$S=\oint ds=\int_0^\theta r\,d\phi=r\theta.$$ 
