• Support PF! Buy your school textbooks, materials and every day products Here!

Improper integral

  • Thread starter trajan22
  • Start date
  • #1
132
0
[tex] \int \frac{dx}{x \sqrt{x^2-4}} [/tex]

there are bounds to this problem but it is irrelevent since my problem is with the integration and not finding the limits.

this integral resembles that of arcsec(x) but im not sure how to deal with the -4.

is there any way to solve this with partial fractions? Or substitution, the squareroot in the denominator is throwing me off for some reason.

sorry for all the recent posts but im trying to teach myself the second part of calculus.

thanks for the help
 
Last edited:

Answers and Replies

  • #2
cristo
Staff Emeritus
Science Advisor
8,107
73
Try the substitution x=2sec(u)
 
  • #3
132
0
ok but why would i pick that as the substitution, what strategy did you use to choose that as a substitution
 
  • #4
cristo
Staff Emeritus
Science Advisor
8,107
73
ok but why would i pick that as the substitution, what strategy did you use to choose that as a substitution
Well, look at the sqrt in the denominator. We know that sec2x-1=tan2x, and that the derivative of secx is secx*tanx. This hints towards the substitution x=secu. Since we want to be able to factor the 4 out of the sqrt (to enable us to use the trig identity above), we then take x=sqrt(4)secu=2secu.
 
  • #5
132
0
Sorry I've been sitting here trying to understand this but I'm still confused. How does this substitution help because if we make x=2secu then we have

[tex] \int \frac{dx}{(2sec(x))\sqrt{(2sec(x))^2-4}} [/tex]

but how is this more manageable than the previous equation.
 
  • #6
1,074
1
Sorry I've been sitting here trying to understand this but I'm still confused. How does this substitution help because if we make x=2secu then we have

[tex] \int \frac{dx}{(2sec(u))\sqrt{(2sec(u))^2-4}} [/tex]

but how is this more manageable than the previous equation.
I changed all your "x"s to "u"s in the integral because that is what they should be after the substitution, but you forgot to substitute for dx after making that substitution. Also it would help to simplify the part under the radical using a trigonometric identity.
 
  • #7
132
0
Right, sorry I was thinking of something else when i did that substitution. I think I understand this now, thanks for all the input.
 

Related Threads on Improper integral

  • Last Post
Replies
7
Views
792
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
715
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
12
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
606
  • Last Post
Replies
1
Views
710
  • Last Post
Replies
13
Views
2K
Top