Integrating Definite Integral: sqrt2 to 2 1/x^3*sqrt(x^2-1)dx

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Homework Help Overview

The discussion revolves around evaluating a definite integral of the function 1/x^3 * sqrt(x^2 - 1) from sqrt(2) to 2. Participants are exploring various methods of integration and substitution techniques.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss separating the integral into two parts and using trigonometric substitution. There are questions about the validity of splitting the integral and the correct substitutions to use. Some suggest using u-substitution to simplify the expression.

Discussion Status

There is an ongoing exploration of different substitution methods, with some participants questioning the approach of splitting the integral. Guidance has been offered regarding the use of specific substitutions, but there is no clear consensus on the best method to proceed.

Contextual Notes

Participants express confusion over the integration process and the resulting expressions, indicating a need for clarification on the relationships between variables in their substitutions.

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Homework Statement


i'm taking a defitinite integral from sqrt2 to 2 of the function 1/x^3*sqrt(x^2-1)dx.

Homework Equations





The Attempt at a Solution


I separated it into 1/x^3 and 1/sqrt(x^2-1). I have the second part using trig sub. as being sec theta dtheta, before integrating it. I believe i did this part correctly.

What I can't remember is that I make 1/x^3 to x^-3 and then integrate it that way with the final being -1/2(1/x^2)??
 
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I know the forums are slow, but you can stop posting this now. The substitution you want is u^2=x^2+1. Work it out. You can turn it into a rational function in u. Then use partial fractions.
 
for some reason this forum isn't working, and i didn't mean to repost several times... I hit refresh a couple times and that's what happened. I apologize!
 
S'ok. Can you remove the other ones or mark as duplicates?
 
Dick said:
The substitution you want is u^2=x^2+1. Work it out. You can turn it into a rational function in u. Then use partial fractions.

Ok, was i correct in splitting it into two integrals?

I'm having a lot of difficulty on this one. I got an answer, but it's really long and seems wrong.

I found I'm supposed to get but need to be able to work through this problem.


How i got mine was for the second part used x = sec y and dx= sec y tan y dy.
after working the substitution i came up with the integral of sec y dy from sqrt(2) to 2. the integration of sec is ln abs(sec y + tan y). I then substituted the inverse sec (x/a) in for y based on my original substitution.

that is then multiplied by the integral of 1/x^3 which is -1/(2x^2). Was this the correct way in doing it?
 
You can't 'split it into two integrals'. There's no rule that says integral(f*g)=integral(f)*integral(g). That's just plain wrong. You can probably handle it with a trig substitution as well, but I would suggest using my original suggestion of u^2=x^2+1.
 
Dick said:
You can't 'split it into two integrals'. There's no rule that says integral(f*g)=integral(f)*integral(g). That's just plain wrong. You can probably handle it with a trig substitution as well, but I would suggest using my original suggestion of u^2=x^2+1.

ok. do you mean to say substitute u^2=x^2 -1 to get rid of the sqrt then?

i just don't understand how that helps me. I end up with variables x and u in the denominator then.
 
Yes, I meant to say u^2=x^2-1. Sorry. It still works. The stray x will cancel out. x^2=u^2+1.
 
Last edited:
Dick said:
Yes, I meant to say u^2=x^2-1. Sorry. It still works. The stray x will cancel out. x^2=u^2-1.

what would x^3 be in terms of u? I'm going to have to head to a tutor and have someone walk me through this one.
 
  • #10
miller8605 said:
what would x^3 be in terms of u?
Here is a hint: don't express x3 in terms of u. At first, express only the square root term and dx in terms of u and du. This will give you an easier term than x^3 to express in terms of u.
 

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