Circular Argument in Proof of Circumference of a Circle using Calculus I have some doubts in this demonstration:

Prove the Circumference of a Circle is $C=2 \pi r$
The equation of a circle is
$(x-h)^2 + (y-k)^2 = r^2 $
To do computations easier let's consider a circle centered in the
origin (0,0)
$x^2 + y^2 = r^2 $
Solving for y we have:
$y = \pm \sqrt{r^2-x^2} $
We know from calculus that we can compute the arc lenght of curves by
using the next formula
$L= \int dL = \int \sqrt{1+(\frac{dy}{dx})^2} dx$
Let's compute the arc length of half a circle ($ L_{hC}  $), to do
that we take the positive solution of the equation of the circle
$y = \sqrt{r^2-x^2}  $
We multiply the final result by two to have the complete answer ($ L_C = 2 L_{hC} $)
So we have
$L_{hC} = \int_{-r}^r  \sqrt{1+(\frac{dy}{dx}^2)}dx ​$
Let's ​compute the derivative of y by doing a change of variable
$u=r^2 -x^2$
$du=-2xdx$
So
$y=\sqrt{u}$
Deriving
$\frac{dy}{du} = \frac{1}{2} (u)^{-\frac{1}{2}}$
$\frac{dy}{dx} = -x(r^2-x^2)^{-\frac{1}{2}}   $
Placing this result
$L_{hC} = \int_{-r}^r \sqrt{1 + (-x(r^2 - x^2)^{- \frac{1}{2} } )^2 }dx $
$L_{hC} = \int_{-r}^r \sqrt{1+ x^2(r^2-x^2)^{-1}} dx$
$L_{hC} = \int_{-r}^r \sqrt{\frac{r^2}{r^2 - x^2}}dx =\int_{-r}^r \frac{r}{\sqrt{r^2 - x^2}} dx$
$L_{hC} =  r  \int_{-r}^r \frac{1}{\sqrt{r^2 - x^2}} dx$
Using trigonometric substitution ($h=r$, $c1=\sqrt{r^2 -x^2}$, $c2=x$)
$ \sin \theta = \frac{x}{r} $
$ x = r \sin \theta $
$ dx = r \cos \theta d \theta $
$ \theta = \arcsin (\frac{x}{r})   $
We also have:
$ \cos \theta = \frac{\sqrt{r^2-x^2}}{r}    $
$ r\cos \theta = \sqrt{r^2 - x^2}   $
Replacing in the integral and changing the integral limits
$ L_{hC} = r \int_{\arcsin(-1)}^{\arcsin(1)} \frac{r\cos \theta}{r\cos \theta} d\theta $
$ L_{hC} = r \int_{- \frac{\pi}{2}}^{ \frac{\pi}{2}} d \theta = r \theta \bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = r(\frac{\pi}{2} - (- \frac{\pi}{2}) ) = r \pi $
Since $ L_C = 2 L_{hC} $ we have $ L_C = 2 \pi r  $
So we prove that the Circumference of a Circle is $C=2 \pi r$

My problem with this demonstration is that it looks like we are using what we are trying to prove in the proof itself.
How do we know that $\arcsin(1)= \frac{\pi}{2}$?
We're using angles in radians. How do we know that the angle of a whole circle is equal to $2\pi$ radians? this piece of information is what we are trying to demonstrate
 A: Before we get into why $\arcsin(1)$ = $\frac{\pi}{2}$ and a full rotation = 360˚ = 2$\pi$ radians, lets look at how radians work:
A angle in radians allows us to form a relationship between the angle, the radii and the arc length (which is harder to do in degrees).
The relationship is something like this:
Circle Image
where the arc length is relative to the angle is:
arc length = angle (in radians) $\times$ radius
$\therefore$ for 1 radian, arc length = radius.
Now to answer your question of: How do we know that the angle of a whole circle is equal to 2 radians?
If according to the proof in your question, where the circumference of the circle, which is equal the arc length of the entire circle, = 2$\pi r$, and our equation,
$\mathbf L$ = $\theta$ * r, where r = radius, $\theta$ = angle in radians and $\mathbf L$ = arc length:
If,
$\mathbf L$ = $\theta$ * r,
$\theta$ = $\frac{\mathbf L}{r}$
But again,
circumference = full arc length = $2\pi r$
$\therefore$ $\theta$ (full) = $\frac{2\pi r}{r}$ = $\mathbf2\mathbf\pi$
The relationship between angle and length is also seen in Trigonometry:
This picture should be pretty much self explanatory
This shows that the sine and cosine of an angle is the ratio of the two sides of a triangle.
Moving to your other question: How do we know that arcsin(1)=  / 2?
To prove this, we first need to prove its inverse: $\sin{(\frac{\pi}{2})} = 1$:
The Unit Circle:
Unit Circle
we can see that:
$\sin(\theta) = y$ (coordinate)
Basically above, we have applied the basic laws of Trigonometry from the right angled triangle, where,
$\sin(\theta) = \frac{opposite}{hypotenuse}$,
but in a unit circle the hypotenuse = radius = 1
So, $\sin(\theta) = \frac{opposite}{hypotenuse}$ = $\frac{opposite}{1}$ = ${opposite}$;
We can tell that $\sin{(\frac{\pi}{2} = 90˚)} = 1$ by inferring; if y is 1 and $\sin(\theta) = opposite$ = y,
$\theta$ = 90˚ counter-clockwise from the x-axis (positive direction from the x-axis).
When $\theta$ = 90˚, on the unit circle, the triangle will look like this... wait... not really a triangle is it?
Visually, a triangle does not exists when two of the angles in a right angle triangle are 90˚ (there will not be a third angle, and a triangle cannot exist without three angles), as the rule of sum of all three angles = 180˚ holds true.
However, by looking at the trend we can tell that as $\theta$ approaches 90˚, $\sin(\theta)$ approaches 1, giving the follow expression:
$\lim_{\theta\to 90˚} \sin{(\theta)} = 1$
(To note: angles above 90˚ repeat the trend (look at the sine graph / curve for more info);
--> The opposite approaches the length of the hypotenuse or radius.
You can also use the sine curve on a cartesian plane to prove this
(where $y = \sin{(x)}$)
Hope this helps :) !!
