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

[tylersmith7690]

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.?

 

[Curious3141]

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.

 

[vanhees71]

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 ;-).