What is the alternate form of $\,\arcsin x\,?$ $\sin x$ can be expressed as 
$$\frac{e^{ix} - e^{-ix}}{2i}$$
through transformations using Euler's formula. I am wondering if $\arcsin x$ has an equivalent, perhaps in logarithms. 
Can we find the inverse of the above equation? 
 A: Notice that
$$x=\sin(\arcsin(x))=\frac{e^{i\arcsin(x)}-e^{-i\arcsin(x)}}{2i}$$
Multiply both sides by $u=e^{i\arcsin(x)}$ to get
$$xu=\frac1{2i}u^2-\frac1{2i}$$
Which is a quadratic in $u$.  Solving for $u$ can be done with the quadratic formula, and then $\arcsin(x)$ may be found with logarithms.
The end result is
$$\arcsin(x)=\frac1i\ln\left(ix\pm\sqrt{1-x^2}\right)$$
Notice the quadratic formula inside the logarithm?
A: Yes, this is how the inverse $\sin$ function is expressed via the $\log$ function:
$$
\arcsin(x) = -i \log(i x + \sqrt{1 - x^2})
$$
Interesting counterparts to this are the following expressions of $\log(x)$:
$$
\log(x) = i \arcsin\left({1 + x^2\over2 x}\right) - {\pi i\over2}  \ \ \ \qquad\mbox{ for }  x\ge1, \qquad
$$
$$
\log(x) = - i \arcsin\left({1 + x^2\over2 x}\right) + {\pi i\over2} \qquad\mbox{ for } 0<x\le1.
$$
A: $y = \sin x = \frac {e^{ix} - e^{-ix}}{2i}$
$x = \arcsin y$
So we just need to solve for $x.$
It will be a little bit easier if we do this substitution.
$e^{ix} = z$
$2iy = z - \frac 1z\\
z^2 - 2iyz -1 = 0$
Apply the quadratic formula
$z= iy \pm \sqrt {1-y^2}$
One solution equals $z$ the other solution equals $\frac 1z$ pick one.  And lets reverse the substitution.
$e^{ix} = iy + \sqrt {1-y^2}\\
x = -i \ln (iy + \sqrt {1-y^2})$ 
A: You can write it as 
$$\arcsin(x) = \int \limits_0^x \frac{dt}{\sqrt{1 - t^2}} $$
which has a certain (far fetched) similarity to the logarithm:
$$\log(x) =  \int_1^x \frac 1tdt$$
A: By definition, $\arcsin x$ is the (one of the) solution(s) of $$\frac{e^{iy}-e^{-iy}}{2i}=x$$ and if $e^{iy}-e^{-iy}=2ix$, then $$ y=\arcsin(x) = -i\log\left(ix\pm\sqrt{1-x^2}\right). $$
