Hyperbolic functions/integration

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  • #1
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Homework Statement


Show that: (I don't have any way of showing it properly so have written it in words)

'The integral from 5 to 3 of (sq root (x^2 - 9) ) with respect to x' = 10 - (9/2) * ln 3

I hope that makes sense!

Homework Equations


Well I'm not sure.. I *think* that the fact the integral of 1/(sq root (x^2 - a^2)) is arcosh (x/a) is relevant.


The Attempt at a Solution


This is part of the 'hyperbolic functions' section on my exam; so the first thing I know is that they will come into it somehow. I cannot solve this by inspection, or by simply comparing it to all the formulae on my sheet.. I also tried to solve by parts (using difference of two squares to give me a root multiplied by another root) but found that this still gave me something I do not know how to integrate. I know if it was 1/(sq root (x^2 - 9)) it would be easy; but cannot work out how the integral of a particular thing and the integral of 1/it are related.

Thanks in advance.
 

Answers and Replies

  • #2
HallsofIvy
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Homework Statement


Show that: (I don't have any way of showing it properly so have written it in words)

'The integral from 5 to 3 of (sq root (x^2 - 9) ) with respect to x' = 10 - (9/2) * ln 3

I hope that makes sense!

Homework Equations


Well I'm not sure.. I *think* that the fact the integral of 1/(sq root (x^2 - a^2)) is arcosh (x/a) is relevant.


The Attempt at a Solution


This is part of the 'hyperbolic functions' section on my exam; so the first thing I know is that they will come into it somehow. I cannot solve this by inspection, or by simply comparing it to all the formulae on my sheet.. I also tried to solve by parts (using difference of two squares to give me a root multiplied by another root) but found that this still gave me something I do not know how to integrate. I know if it was 1/(sq root (x^2 - 9)) it would be easy; but cannot work out how the integral of a particular thing and the integral of 1/it are related.

Thanks in advance.
It is true that [itex]cosh^2(\theta)- sinh^2(\theta)= 1[/itex] so, dividing both sides by [itex]cosh^(\theta)[/itex], [itex]1- tanh^2(\theta)= csch^2(\theta)[/itex] . That is, if you let [itex]x= 3coth(\theta), [itex]9- x^2= 9- 9coth^2(\theta)= 9csch^2(\theta)[/itex] so [itex]\sqrt{9- x^2}= \sqrt{9csch^2(\theta)}= 3csch(\theta)[/itex]. Of course, [itex]3(coth(\theta))'= 3(1- coth(\theta))[/itex] so [itex]dx= 3(1- coth(\theta))d\theta[/itex].
The integral becomes
[tex]9\int coth(\theta)(1- coth(\theta)d\theta[/tex]
 
  • #3
malty
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Don't think arccos is relevant. . .

Hmm I don't think a hyperbolic substition is necessary here . ..need to check my calculations ... .:eek:

//Argh forgot to square the minus// Nevermind::redface:
 
Last edited:
  • #4
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Right. eek.. I really didn't follow that! sorry. right..

what does [itex] mean?? and how did you get from sq root(x^2 - 9) to sq root (9 - x^2)? without getting into complex numbers..
 
  • #5
malty
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Right. eek.. I really didn't follow that! sorry. right..

what does i tek mean?? and how did you get from sq root(x^2 - 9) to sq root (9 - x^2)? without getting into complex numbers..



I dunno how he did that, because that's what's worrying me each time I do this integral I get a complex function ... *not having a good day:(*

[ ite k or latek" is just a thing for typing lovely equations, e.g [tex]\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
 
Last edited:
  • #6
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Well I tried using the substitution x = 3coth theta... though I'll call theta y to make it easier to type.
I then get x^2 - 9 = 9coth^2 y - 9
root (x^2 - 9) = 3(cosech y)

therefore the original question is:
the integral from x=5 to x=3 of 1/(3 (cosech y)) dx.

dx/dy = -3(cosech y) ... I think, not so sure on my maths there. now I get stuck again as this seems to cancel to give no variable to integrate..
 
  • #7
malty
Gold Member
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Well I tried using the substitution x = 3coth theta... though I'll call theta y to make it easier to type.
I then get x^2 - 9 = 9coth^2 y - 9
root (x^2 - 9) = 3(cosech y)

therefore the original question is:
the integral from x=5 to x=3 of 1/(3 (cosech y)) dx.

dx/dy = -3(cosech y) ... I think, not so sure on my maths there. now I get stuck again as this seems to cancel to give no variable to integrate..

...oopps... never mind that

[tex] \frac{dx}{d\theta}=3 cosch^2 ({\theta}) [/tex] .. .



Starting over grrr


Yay i've finally got it!! ... I think
 
Last edited:
  • #8
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I found a solution online (the cheats way out I know!).. but I'll do it properly when I get time. Its easier to use a different substitution (x = 3 cosh y, if I remember rightly).
Thanks anyway.
 
  • #9
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sorry HallsofIvy : 1- tanh^2 (theta)= sech^2(theta) !!
thanks
 

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