Geometric intuition for expressing $\arcsin(x)$ as an integral? For $|x| \leq 1$ we have that:
$$\arcsin(x)=\int_0^ x\frac1{\sqrt{1-z^2}}~\mathrm dz$$
Okay... So we are integrating the reciprocal of a function that would parameterize a half of the unit circle, and $\arcsin(x)$ is the inverse of the restriction of the $\sin$ function to a half of the unit circle. If the tangent line to the graph of $f(x)$ at $x=a$ is equal to $m$ then the tangent line to the graph of $f^{-1}(x)$ at $x=f^{-1}(f(a))$ is $\frac{1}{m}$. I can't help but feel that there is some geometric insight happening here that would make this $\arcsin(x)$ identity a whole lot more transparent...
 A: This follows from the arc length formula $L=\int_a^b\sqrt{1+(dy/dx)^2}$. To obtain $\arcsin y$, we imagine $y$ as a height on the vertical axis. To find arcsine is to find the angle $\theta$ such that $y$ is the height corresponding to that angle:

The arclength corresponding to an angle is the same as the angle itself on the unit circle. Since $x^2+y^2=1$, we have $y=\sqrt{1-x^2}$, and so
$$\int_0^y\sqrt{1+\Bigl({dy\over dx}\Bigr)^2}\,dx
=\int_0^y\sqrt{1+{x^2\over1-x^2}}\,dx
=\int_0^y{dx\over\sqrt{1-x^2}}.$$
A: There's clearly a nice geometric story here, and I'm curious what people come up with. Too long for a comment, but here's a derivation of the identity above. I'll be loose about restrictions on $x$ because I'm too lazy to think about signs and so on.
We can start from
$$ \sin(\alpha) = x$$
and wonder, what's the angle $\alpha(x)$?
We know the Pythagorean theorem holds:
$$\cos^2(\alpha)+\sin^2(\alpha)=1, $$
So that
$$ \cos(\alpha) = \sqrt{1-x^2}.$$
Taking a derivative of this wrt $x$ provides
$$ \alpha'(x) = \frac{1}{\sqrt{1-x^2}}, $$
from which it follows that our angle is
$$\alpha(x) = \int_0^x \frac{dz}{\sqrt{1-z^2}},$$
meaning that
$$ \sin\Big(\int_0^x \frac{dz}{\sqrt{1-z^2}} \Big) = x.$$
So the identity is really just a direct consequence of the Pythagorean theorem. Anything beyond this is likely to be an alternate way of seeing the same thing, perhaps reasoning with circles or squares instead of triangles.
