Picture Justification of $\arccos(x) = \int_x^1 \frac{1}{\sqrt{1-t^2}}dt$ Is there a quick shortcut way to write down that
$$\arccos(x) = \int_x^1 \frac{1}{\sqrt{1-t^2}}dt$$
just using the picture here?
I know it can be computed with a computation using $\theta = \frac{s}{r}$ and 
$$\theta = \frac{s}{1} = \int_0^s ds' = \int_x^1 \sqrt{1+y'}dt = \int_x^1 \sqrt{1+\left(\frac{d}{dt}\sqrt{1-t^2}\right)^2}dt = ... = \int_x^1 \frac{1}{\sqrt{1-t^2}}dt$$
but it should be obvious without any calculation right? Don't see / forget how to justify it. The main issue is justifying the 
$$\frac{1}{\sqrt{1-t^2}}$$
term.
 A: Heuristic Development:
I'm not sure if there is a quick way to derive the integral representation for the arccosine function.  But here is a way we can proceed using the "picture" and applying heuristic reasoning.
Draw an incremental segment of length $ds$ at a point on the circle at angle $s$; this segment is tangent to the circle. Then draw a small right triangle with sides $dx$ and $dy$, parallel to the $x$-axis and $y$-axis, respectively.  Denote the angle that $dx$ makes with $ds$ by $\gamma$.  
Then, note that the differential arc length $ds$ can be expressed as a projection onto the $x$-axis as
$$ds=\frac{dx}{\cos \gamma}$$
But, we can easily see that $\cos \gamma =\sin (s) =\sqrt{1-x^2}$.  And we have 
$$ds=\frac{dx}{\sqrt{1-x^2}}$$

Rigorous Approach:
I thought it might be instructive to present a purely analytical approach (i.e., one that needs no "picture.") that relies only on knowledge of the cosine function and its derivative.
We have
$$f(x) = \cos (x)\implies f^{-1}(x)=\arccos (x)$$
Then, using THIS relationship of the derivative of an inverse function we have
$$f'(x)=-\sin x=-\sqrt{1-\cos^2x}\implies (f^{-1})'(x)=-\frac{1}{\sqrt{1-x^2}} \tag 1$$
Now, integrating $(1)$ we obtain
$$\begin{align}
\int_x^1 \frac{1}{\sqrt{1-t^2}}\,dt&=-\int_x^1 (f^{-1})'(t)\,dt\\\\
&=-\int_x^1 \frac{d\arccos(t)}{dt}\,dt\\\\
&=\arccos(x)-\arccos(1)\\\\
&=\arccos(x)
\end{align}$$
And of course, we could work backwards by starting with the arccosine written as
$$\arccos(x)=-\int_x^1 \frac{d\arccos(t)}{dt}\,dt$$
Then, using $(1)$, we arrive at the coveted result.

NOTE:
If we use the notion of arc length to derive the arccosine, then we proceed as follows.  
The equation for the upper-half of the circle is given by $y(x)=\sqrt{1-x^2}$, for $-1< x<1$.  Then, the arc length for an arc that subtends $s$ can be written
$$\begin{align}
s&=\int_0^s ds'\\\\
&=\int_{\cos s}^1 \sqrt{1+\left(\frac{dy}{dx'}\right)^2}\,dx'\\\\
&=\int_{\cos s}^1 \frac{1}{\sqrt{1-x'^2}}\,dx' \tag 1
\end{align}$$
Then, letting $s=\arccos(x)$ in $(1)$ reveals that
$$\arccos(x)=\int_x^1 \frac{1}{\sqrt{1-x'^2}}\,dx'$$
