- #1
PFuser1232
- 479
- 20
What exactly is the integral of ##\frac{1}{\sqrt{x^2 - 1}}##?
I know that the derivative of ##\cosh^{-1}{x}## is ##\frac{1}{\sqrt{x^2 - 1}}##, but ##\cosh^{-1}{x}## is only defined for ##x \geq 1##, whereas ##\frac{1}{\sqrt{x^2 - 1}}## is defined for all ##|x| \geq 1##. How do I take that into account? Do I write:
$$\int \frac{1}{\sqrt{x^2 - 1}} dx = \cosh^{-1}{|x|} + c$$
UPDATE: I tried differentiating ##\cosh^{-1}{|x|}## as a piecewise function and I ended up with ##-\frac{1}{\sqrt{x^2 - 1}}## for ##x \leq 1##. Now I'm more confused.
Should I write the integral as:
$$\int \frac{1}{\sqrt{x^2 - 1}} dx = S(x) \cosh^{-1}{|x|}$$
where ##S## is the signum function?
I know that the derivative of ##\cosh^{-1}{x}## is ##\frac{1}{\sqrt{x^2 - 1}}##, but ##\cosh^{-1}{x}## is only defined for ##x \geq 1##, whereas ##\frac{1}{\sqrt{x^2 - 1}}## is defined for all ##|x| \geq 1##. How do I take that into account? Do I write:
$$\int \frac{1}{\sqrt{x^2 - 1}} dx = \cosh^{-1}{|x|} + c$$
UPDATE: I tried differentiating ##\cosh^{-1}{|x|}## as a piecewise function and I ended up with ##-\frac{1}{\sqrt{x^2 - 1}}## for ##x \leq 1##. Now I'm more confused.
Should I write the integral as:
$$\int \frac{1}{\sqrt{x^2 - 1}} dx = S(x) \cosh^{-1}{|x|}$$
where ##S## is the signum function?
Last edited: