I recently found a different approach to $\int{\sec(x)}dx$ as opposed to the common solution of $\frac{1}{2}\ln{\frac{1+\sin(x)}{1-\sin(x)}}+C$: (I'm sure it exists somewhere else and I've just been unable to find it-so any help on that front would be greatly appreciated)
$$\int{\sec(x)}dx = \int\frac{1}{\tan(x)}\frac{d}{dx}(\sec(x))dx$$ since $\frac{d}{dx}(\sec(x))$ is $\sec(x)\cdot \tan(x)$, furthermore
$$=\int\frac{1}{\sqrt{(\sec(x))^2-1}}d\sec(x)=\int\frac{1}{\sqrt{u^2-1}}du$$
Then, $$\int\frac{1}{i\sqrt{u^2-1}}du=i\cdot\arcsin(u)+C=i\cdot\arcsin(\sec(x))+C$$
How would one prove that this complex solution is equal to the more common solution above without just taking the derivative? The only potentially relevant thing I've managed to think of is Euler's formula: $e^{ix}=\cos(x)+i\cdot \sin(x)$
Have you ever worked out $\displaystyle\int \frac{dx}{x^2+1}$ using partial fractions and related $\arctan x$ and $\frac i2\log\big(\frac{x-i}{x+i}\big)$?
You have a sign error, however, in your presentation. You want $\displaystyle{\int\frac{du}{\sqrt{u^2-1}} = i\int\frac{du}{\sqrt{1-u^2}}}.$
Set $\theta = i\arcsin(\sec x)$, so $\sec x = \sin(-i\theta) = \dfrac{e^\theta-e^{-\theta}}2$. Thus, solving the quadratic equation, we have $$e^\theta = \frac12\left(2\sec x \pm \sqrt{4\sec^2x - 4}\right) = \sec x \pm \sqrt{\sec^2x-1} = \sec x\pm \tan x.$$ I'll let you finish.