I understand the similarities of a sine function and a hyperbolic sine function. They are both sums of natural exponential function but the difference is that one is complex and the other is real. In other words $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2}$$ and $$\sinh(x)=\frac{e^{x}-e^{-x}}{2}$$
My question is if I was to find the anti derivative of $\frac{1}{\sqrt{1-x^{2}}}$. Here is what I would do.
Let $x=\sin(t) \Rightarrow dx=\cos(t)\,dt$. $$\int\frac{dx}{\sqrt{1-x^{2}}}=\int\frac{\cos(t)dt}{\sqrt{1-\sin(t)^{2}}}$$ $$=\int{dt}=t=\sin^{-1}(x)$$
The assumption that $x=\sin(t)$ has created a solution as shown above. But if I were to start with a new assumption that $x=\sinh(t)$. By following the same procedure that the solution to the anti derivative becomes $\sinh^{-1}(x)$. Now this shows that the two solutions are possible for the anti derivative of $\frac{1}{\sqrt{1-x^{2}}}$. Now if we make this practical and look at a sag of transmission line, it follows a hyperbolic function derived from the above anti derivative. But if we started with a sine function we would have concluded that the sag of a transmission line is a sinusoid (which is not the case). So doesn't this create a contradiction.
You are wrong. The procedure that you applied in order to find an antiderivative of $\frac1{\sqrt{1-x^2}}$ is correct. But it doesn't work if you use $\sinh$ instead of $\sin$, because $1-\sinh^2\neq\cosh^2$. The similar equality that holds is $1+\sinh^2=\cosh^2$. And you can use it to find an antiderivative of $\frac1{\sqrt{1+x^2}}$.