# Integral of $\frac{1}{\sqrt{x^2-1}}$: Solving & Understanding

• PFuser1232
In summary, the integral of the function ##\frac{1}{\sqrt{x^2 - 1}}## is the same as the inverse hyperbolic cosine.
PFuser1232
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?

Last edited:
Consider a comparison to ln(x)

notice how:
##\frac{d}{dx}\ln(x)## is the same as ##\frac{d}{dx}\ln(-x)##

This is inconsistent with your ##\cosh^{-1}{|x|}##

The idea that resolves this is that you can use the idea that ##\frac{1}{\sqrt{x^2 - 1}}## is an even function.

There could be some quirks using complex numbers... I'll try it and get back.

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?

Last edited:
PFuser1232
one idea is to consider ##\cosh(x) = cos(ix)##

i.e.
##\cosh^{-1}(x) = -i cos^{-1}(x)##

Most obvious solution would be your last line

Last edited:
Stephen Hodgson said:
one idea is to consider ##\cosh(x) = cos(ix)##

i.e.
##\cosh^{-1}(x) = -i cos^{-1}(x)##

Most obvious solution would be your last line
Stephen Hodgson said:
Consider a comparison to ln(x)

notice how:
##\frac{d}{dx}\ln(x)## is the same as ##\frac{d}{dx}\ln(-x)##

This is inconsistent with your ##\cosh^{-1}{|x|}##

The idea that resolves this is that you can use the idea that ##\frac{1}{\sqrt{x^2 - 1}}## is an even function.

There could be some quirks using complex numbers... I'll try it and get back.

I think I figured it out. It's ##\ln|x + \sqrt{x^2 - 1}|##.
Which is identical to the inverse hyperbolic cosine, except it's defined for both positive and negative ##x + \sqrt{x^2 - 1}##.

yeah, that's a much nicer way of writing it.

## 1. What is the integral of $\frac{1}{\sqrt{x^2-1}}$?

The integral of $\frac{1}{\sqrt{x^2-1}}$ is $\ln|x+\sqrt{x^2-1}|+C$, where C is the constant of integration.

## 2. What is the process for solving the integral of $\frac{1}{\sqrt{x^2-1}}$?

The process for solving the integral of $\frac{1}{\sqrt{x^2-1}}$ is to first rewrite the integrand using the substitution $u=x^2-1$. Then, use the power rule for integration to find the antiderivative, and finally substitute back in the original variable.

## 3. Why is it important to understand the integral of $\frac{1}{\sqrt{x^2-1}}$?

Understanding the integral of $\frac{1}{\sqrt{x^2-1}}$ is important because it is a common integral that arises in many different areas of mathematics and science. It is also a good example for practicing integration techniques and developing a deeper understanding of calculus.

## 4. Are there any specific applications for the integral of $\frac{1}{\sqrt{x^2-1}}$?

Yes, the integral of $\frac{1}{\sqrt{x^2-1}}$ has several applications in physics and engineering, such as calculating work done by a force or finding the center of mass of a curved shape.

## 5. What is the relationship between the integral of $\frac{1}{\sqrt{x^2-1}}$ and the inverse trigonometric functions?

The integral of $\frac{1}{\sqrt{x^2-1}}$ can be rewritten as $\ln|x+\sqrt{x^2-1}|+C=\cosh^{-1}(x)+C$, where $\cosh^{-1}(x)$ is the inverse hyperbolic cosine function. This relationship can be useful in solving more complex integrals involving inverse trigonometric functions.

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