Using complex number in integration I know,
$$I_1 = \int \dfrac{dx}{\sqrt{x^2 - 1}} = \ln|x + \sqrt{x^2-1}| + c$$
But if I factor out $i$ from the denominator, I get:
$$I_2 = -i \int \dfrac{dx}{\sqrt{1 - x^2}} = -i \sin^{-1}x + ic$$
Are these 2 expressions equivalent?

Let $\sin^{-1}x = θ$
$$ \sin θ = x\\
\frac{e^{iθ} - e^{-iθ}}{2i} = x \\ $$
Solving the quadratic,
$$ e^{iθ} = ix \mp \sqrt{1 - x^2} = i(x \mp \sqrt{x^2 - 1})\\ $$
Taking log on both sides and then multiplying the numerator and denominator by the congugate,
$$\begin{align}
iθ &= \ln i + \ln |x \mp \sqrt{x^2 - 1}| \\
i \sin^{-1}x &= \ln e^{\frac{iπ}2} + \ln |\frac 1{x \pm \sqrt{x^2 - 1}}|\\
&= \frac{iπ}2 - \ln |x \pm \sqrt{x^2 - 1}|\\
\end{align}$$
So $ - i \sin^{-1}x$ is equivalent to $\ln |x ± \sqrt{x^2 - 1}|$.
