I was dealing with the following differential equation in the context of catenaries:
$$\frac{dy^\prime} {dx}=\sqrt{1+(y^\prime)^2}$$
Separation of variables gives
$$\int{ \frac{d y^\prime}{\sqrt{1+(y^\prime)^2}} } = \int{dx} = x + C$$
The derivation I was shown then proceeds to reason thus: If we have an integral in the form
$$ \int{ \frac{du}{\sqrt{1-u^2}} } $$
then the trigonometric substitution $u=\sin{t}$, with $-\frac{\pi}{2} \lt t \lt \frac{\pi}{2}$ gives
$$ \int{ \frac{du}{\sqrt{1-u^2}} }= \int{ \frac{\cos{t} \ dt} { \sqrt{1-\sin^2{t}} }}= \int{ \frac{\cos{t} \ dt} {\cos{t}} }= \int{dt}=t+C= \arcsin{u}+C $$
We can then take our original integral and write it as
$$\int{ \frac{d y^\prime}{\sqrt{1-(iy^\prime)^2}} }$$
If we set $iy^\prime=\sin{t}$, $-\frac{\pi}{2} \lt t \lt \frac{\pi}{2}$, then $dy^\prime=\frac{\cos{t}}{i} \ dt$ and we get
$$\int{ \frac{dy^\prime}{\sqrt{1-(iy^\prime)^2}} }= \frac{1}{i}\int{ \frac{\cos{t} \ dt}{\sqrt{1-\sin^2{t}}} }= \frac{1}{i} \arcsin{(iy^\prime)} + C $$
Thus we have
$$ \frac{1}{i} \arcsin{(iy^\prime)}= x+C $$
and therefore
$$ y^\prime= \frac{1}{i} \sin{(i(x+C))} $$
Finally, since we have
$$ \sinh{\theta}=\frac{1}{i}\sin{i\theta} $$
we arrive at
$$ y^\prime = \sinh(x+C) $$
and this is the final solution.
There’s something I don’t quite understand. You could take any $y^\prime \in \mathbb{R}$ and the function
$$ \frac{1}{\sqrt{1-(iy^\prime)^2}} $$
should yield a real number and be continuous. This suggests to me that there is at least some interval in the reals for which the integral exists and that there are at least some $y^\prime$ that are real.
However, if we look at the substitution $iy^\prime=\sin{t}$, $-\frac{\pi}{2} \lt t \lt \frac{\pi}{2}$, $\sin{t}$ is a real number for all $t$ within the given interval. $y^\prime$ is therefore imaginary, since it is equal to $\frac{\sin{t}}{i}$. And yet we know that there are some $y^\prime$ that are not imaginary. Is this substitution valid?