I have found this expression (I am worried about domain issues or any algebraic mistake I make)by playing around with the formula for sine, i.e-
$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$ From this, I got->
$$\sin^{-1}(x)=-i\ln|ix \pm \sqrt{1-x^2}|$$
I think I know why there's a plus minus, I propose the following,
We know, $\sin^{-1}(\sin(x))=x$ so plugging sine into formula we get,
$$=-i\ln|i\sin(x)\pm\sqrt{1-\sin^2(x)}|$$
But since, $\sqrt{1-\sin^2(x)}=|\cos(x)|$ We can split this into cases,
Case (i) $|\cos(x)|=\cos(x)$ Now we want a "+" in the formula so that we can get $e^{ix}=\cos(x)+i\sin(x)$ So we get,
$\sin^{-1}(\sin(x))=-i\ln|\cos(x)+i\sin(x)|$
Simplifying this we get,
$\sin^{-1}(\sin(x))=x$
The formula works!
Now similarly we have a "-" when $|\cos(x)|=-\cos(x)$ for case 2.
Conclusion:Is the following true?
$\sin^{-1}(x)=-i\ln|ix + \sqrt{1-x^2}|$ when $|\cos(x)|=\cos(x)$ and,
$\sin^{-1}(x)=-i\ln|ix - \sqrt{1-x^2}|$ when $|\cos(x)|=-\cos(x)$
In the formal definitions we have
$$ \sin^{-1}(x)=-i\sinh^{-1}(ix)\\ \sinh^{-1}(x)=\ln(x+\sqrt{x^2+1}\big) $$
and thus
$$ \sin^{-1}(x)=-i\ln(ix+\sqrt{1-x^2}\big) $$
Notice that there is no $\pm$ here. This is not an oversight. There are $\pm$ in the definitions of $\cosh^{-1}(x)$ as well as other inverse hyperbolic functions. In addition, you suggest that $|\cos(x)|$ can have positive and negative values. That would not be possible.