# Solve Indefinite Integral: ∫1/sqrt(x2-1)^5 dx

• tylersmith7690
In summary, the student is stuck and is looking for a solution. They are using a trigonometric substitution and evaluating an indefinite integral. They are trying to find a solution using the cosh-function and the sinh-function. They are then trying to integrate using a clever trick involving the cosh-function and the sinh-function. They are getting close to a solution, but they are not sure if it is correct.

## Homework Statement

Sorry for the poor use of Latex, I have tried to get it to work but it seems to never come out as I would like.

Using a trigonometric or hyperbolic substitution, evaluate the following indfe nite integral,

∫$\frac{1}{\sqrt{(x^2-1)^5}}$ dx

## Homework Equations

I have got down to a point where I am stuck and was wondering which path to go down next.

## The Attempt at a Solution

let x=cosh$\phi$
dx/d$\phi$=sinh$\phi$
dx=sinh$\phi$ d$\phi$

Then (x2-1) = cosh2$\phi$-1
= sinh2$\phi$

(x2-1)1/2= sinh$\phi$
(x2-1)5/2= sinh5$\phi$

therefore ∫sinh$\phi$/sinh5$\phi$
=∫1/sinh4$\phi$
=∫cosec2$\phi$

Is this right so far, Do i then split the (cosec$\phi$)^4 into two and do the integral then.?

tylersmith7690 said:

## Homework Statement

Sorry for the poor use of Latex, I have tried to get it to work but it seems to never come out as I would like.

Using a trigonometric or hyperbolic substitution, evaluate the following indfenite integral,

∫$\frac{1}{\sqrt{(x^2-1)^5}}$ dx

## Homework Equations

I have got down to a point where I am stuck and was wondering which path to go down next.

## The Attempt at a Solution

let x=cosh$\phi$
dx/d$\phi$=sinh$\phi$
dx=sinh$\phi$ d$\phi$

Then (x2-1) = cosh2$\phi$-1
= sinh2$\phi$

(x2-1)1/2= sinh$\phi$
(x2-1)5/2= sinh5$\phi$

therefore ∫sinh$\phi$/sinh5$\phi$
=∫1/sinh4$\phi$
=∫cosec2$\phi$

Is this right so far, Do i then split the (cosec$\phi$)^4 into two and do the integral then.?

The last integral should be ##\int (\mathrm{csch} \phi)^4 d\phi##, shouldn't it?

There might well be an easier way to go about this with hyperbolic trig identities, but I would do it with a mainly algebraic integral.

First convert to the exponential form. I'm going to be using y instead of ##\phi## for ease of Latex:

You have to evaluate ##\int \frac{16}{(e^y - e^{-y})^4}dy##.

Sub ##u = e^y##. Simplify. You'll end up with a ##u^3## in the numerator, and a ##(u^2 - 1)^4## in the denominator. There's a clever way to split this up so you can integrate by parts. And this can be repeated one more time.

So far it looks good (in the last line it must be $\text{cosech}^4 x$ of course.

My next guess is that it might help to use
$$\cosh^2 x-\sinh^2 x=1$$
in the numerator of the integrand ;-).