Q:

Is it possible to deduce what is $\int_1^x \sqrt{u^2-1}du$ from $\int_0^x \sqrt{1-u^2}du$?

I was wondering if it was possible to quickly deduce what is $$F(x) =\int_1^x \sqrt{u^2-1}du, x \geq 1$$ from $$G(x) = \int_0^x \sqrt{1-u^2}du = \frac{1}{2}\arccos(x)-\frac{1}{2}x\sqrt{1-x^2} + c, 0 \leq x \leq 1,$$
where $G(x)$ can easily be found using subsitution with a cosine function.
I made a wrongful argument where I just wrote $\sqrt{u^2-1} = i \sqrt{1-u^2}$ and didn't bother about the intervals, so I ended up with a real number equal to a complex number ....

A:

For
$$
1-u^2
$$
It makes sense to use the following identity
$$
\sin^2 x + \cos^2x =1 \implies 1-\sin^2x = \cos^2x
$$
(or vice versa)
For 
$$
u^2-1
$$
we can use the following idenity
$$
\cosh^2x - \sinh^2x = 1 \implies \cosh^2x - 1 = \sinh^2 x
$$
which makes sense to use that sub. which is pretty quick to integrate.

