How to show $\arcsin{x} = \frac{\pi}{2} + i \ln{(x+\sqrt{x^2-1})}$? Is the following identity correct
$\arcsin{x} = \frac{\pi}{2} + i \ln{(x+\sqrt{x^2-1})}?$
Here, $x < 1$. How can we show that it is true? One way to see it is by differentiating, since
$\frac{d}{dx} (LHS) = \frac{d}{dx} (RHS).$ 
Thanks.

One way to see why it is correct (but not to derive it) is to plug back into the definition and grind it through. Recall $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}.$$
Lets plug in $x=\frac{\pi}{2}+i\ln\left(z+\sqrt{z^{2}-1}\right).$ Then we have $$\sin(x)=\frac{1}{2i}\left(\exp\left(\frac{\pi i}{2}+-\ln\left(z+\sqrt{z^{2}-1}\right)\right)-\exp\left(\frac{-i\pi}{2}+\ln\left(z+\sqrt{z^{2}-1}\right)\right)\right)$$
$$=\frac{1}{2}\left(\exp\left(-\ln\left(z+\sqrt{z^{2}-1}\right)\right)+\exp\left(\frac{-i\pi}{2}+\ln\left(z+\sqrt{z^{2}-1}\right)\right)\right)$$
$$=\frac{1}{2}\left(\frac{1}{z+\sqrt{z^{2}-1}}+z+\sqrt{z^{2}-1}\right)=z.$$
The last equality follows from rationalizing the denominator, and the cancellations that follow.
Hope that helps,
