Indefinite Integral

  • #1

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,

∫[itex]\frac{1}{\sqrt{(x^2-1)^5}}[/itex] 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[itex]\phi[/itex]
dx/d[itex]\phi[/itex]=sinh[itex]\phi[/itex]
dx=sinh[itex]\phi[/itex] d[itex]\phi[/itex]

Then (x2-1) = cosh2[itex]\phi[/itex]-1
= sinh2[itex]\phi[/itex]

(x2-1)1/2= sinh[itex]\phi[/itex]
(x2-1)5/2= sinh5[itex]\phi[/itex]

therefore ∫sinh[itex]\phi[/itex]/sinh5[itex]\phi[/itex]
=∫1/sinh4[itex]\phi[/itex]
=∫cosec2[itex]\phi[/itex]

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

Answers and Replies

  • #2
Curious3141
Homework Helper
2,843
87

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,

∫[itex]\frac{1}{\sqrt{(x^2-1)^5}}[/itex] 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[itex]\phi[/itex]
dx/d[itex]\phi[/itex]=sinh[itex]\phi[/itex]
dx=sinh[itex]\phi[/itex] d[itex]\phi[/itex]

Then (x2-1) = cosh2[itex]\phi[/itex]-1
= sinh2[itex]\phi[/itex]

(x2-1)1/2= sinh[itex]\phi[/itex]
(x2-1)5/2= sinh5[itex]\phi[/itex]

therefore ∫sinh[itex]\phi[/itex]/sinh5[itex]\phi[/itex]
=∫1/sinh4[itex]\phi[/itex]
=∫cosec2[itex]\phi[/itex]

Is this right so far, Do i then split the (cosec[itex]\phi[/itex])^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.
 
  • #3
vanhees71
Science Advisor
Insights Author
Gold Member
17,026
8,153
So far it looks good (in the last line it must be [itex]\text{cosech}^4 x[/itex] of course.

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

Related Threads on Indefinite Integral

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
967
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
1
Views
993
  • Last Post
Replies
1
Views
1K
Top