It seems to be an elliptic integral of the second kind, but when $k=i$? This is going by the definition that $E(\theta,k)=\int_{0}^{\theta} \sqrt{1-k^2 \sin^2x}dx$. That seems a bit off.
Or is this not one at all due to the indefinite nature of the integral?
Is $\displaystyle\int\sqrt{1+\sin^2x}~dx~$ an elliptic integral?
Yes, it is. In fact, that's how elliptic integrals came into being, by trying to compute the arc length of the sine or cosine function, which is precisely what you have done here.
But when $k=i$? That seems a bit off.
It does, doesn't it? But why is that such a shock for you? :-) Even trigonometric functions can be written as exponential and/or logarithmic expressions containing imaginary arguments, so why would this be any different?
Consider the elliptic integral of the second kind $E(\phi,m):=\int\limits _0^\theta \sqrt{1-m \sin^2 x}\,dx$. (Note that my convention will be to write in terms of $m=k^2$ rather than $k$ itself.) Naively, the range of allowed $m$ is $m\in [0,1]$. The question posted above is then a special case of the following: What is the meaning of $E(\theta,-m)$ for $0\leq m \leq 1$?
To answer this, We can manipulate $E(\theta,-m)$ into a more recognizable form. Substituting $u=\pi/2-x$, we have $$1+m \sin^2 x =1+m(1-\sin^2 u)=(1+m)\left(1-\frac{m}{1+m}\sin^2 u\right)$$ and therefore
\begin{align} E(\theta,-m) &=(1+m)^{1/2}\int\limits _{\pi/2-\theta}^{\pi/2} \sqrt{1-\frac{m}{1+m}\sin^2 u}\,du \\ &=(1+m)^{1/2} E\left(\frac{\pi}{2},\frac{m}{1+m}\right)-(1+m)^{1/2} E\left(\frac{\pi}{2}-\theta,\frac{m}{1+m}\right). \end{align}
Note that the function in the first term is in fact the complete elliptic integral $E\left(\frac{m}{m+1}\right).$ For the special case of $m=1$, we conclude that
\begin{align} E(\theta,-1) &=\int_{0}^\theta \sqrt{1+\sin^2 x}\,dx \\ &= \sqrt{2}E\left(\frac{1}{2}\right)-\sqrt{2}E\left(\frac{\pi}{2}-\theta,\frac{1}{2}\right)=\sqrt{2}\int_{\pi/2-\theta}^{\pi/2} \sqrt{1-\frac{1}{2}\sin^2 x}\,dx. \end{align}
