Seemingly easy integral

  • Thread starter xAxis
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  • #1
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I came to this integral which I thaught would be easy to solve, but I'm stuck.

int dx/{x *sqrt{(x ^ 2 )+ k}}
(Sorry Im trying to get it in Latex)

I tried substitution:
[tex]x ^ 2 + k = t[/tex]
as I remembered from the book (years ago)

But have ended with this:
1/2 times
int dt/{(t-k)sqrt(t)}

Im not sure if this leads to a solution. Any help would be appreciated
 
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Answers and Replies

  • #2
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Try the substitution x=k1/2tan(y)
 
  • #3
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Thanks d leet. Sorry I tried in Latex and kept getting most unusual expressions. This is the correct expression.
 
  • #4
dextercioby
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Try the substitution x=k^{1/2} sinh t.

Daniel.
 
  • #5
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that substitution led me to
int dt/sinh t
Any idea how this could be solved?
 
  • #6
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xAxis said:
that substitution led me to
int dt/sinh t
Any idea how this could be solved?

It shouldn't lead you to that, can you show the work you did using that substitution?
 
  • #7
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Sure if you don't mind my ascii typing :)
dx = (sqrt)k cosht dt
I = (sqrt)k*int cosht*dt/((sqrt)k*sinht*(sqrt)(k(sinh^2t + 1))) =

=(sqrt)k*int cosht*dt/((sqrt)k*sinht*(sqrt)(k*cosh^2t)) =

==(sqrt)k*int cosht*dt/(k*sinht*cosht)

Then after canceling cosht and puting 1/k out, Im left with dt/sinht
 
  • #8
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xAxis said:
Sure if you don't mind my ascii typing :)
dx = (sqrt)k cosht dt
I = (sqrt)k*int cosht*dt/((sqrt)k*sinht*(sqrt)(k(sinh^2t + 1))) =

This line is almost impossible to follow, but it doesn't look correct at all. you should have k.5*cosht/(ksinh2 + k).5 inside of the integral.
 
  • #9
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I got the same under the root of the denominator. When you factor out k, u got (k(sinh^2 + 1))^.5 But that's only the root. There was an x multilplying that root.
I think I reduced it properly, but don't know how to solve dt/sinht
 
  • #10
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xAxis said:
I got the same under the root of the denominator. When you factor out k, u got (k(sinh^2 + 1))^.5 But that's only the root. There was an x multilplying that root.
I think I reduced it properly, but don't know how to solve dt/sinht

Sorry about that I didn't see the factor of x in the denominator. Well isn't 1/sech(t)= csch(t)? and I think you can look that up in a table of integrals, or I;m sure theres a way to do it, but I'm not sure how do it since I never have before.
 
  • #11
dextercioby
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that substitution led me to
int dt/sinh t
Any idea how this could be solved?

Yes, just multiply both the denominator and the numerator by #sinh t# and then use the ch^2 -sh^2 =1 identity.

Daniel.
 
  • #12
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Thanks dextercioby, your first tip was good as I managed to reduce to the above. As I didn't find it in tables, I used an excelent web integrator
http://integrals.wolfram.com/index.jsp
The solution is ln(tanh(t/2))
 
  • #13
dextercioby
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There's also the substitution

[tex] x^{2}+1 =p^{2} [/tex]

that leads immediately to a result.

Daniel.
 
  • #14
benorin
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Thanks dextercioby, your first tip was good as I managed to reduce to the above. As I didn't find it in tables, I used an excelent web integrator
http://integrals.wolfram.com/index.jsp
The solution is ln(tanh(t/2))

www.quickmath.com is an excellent alternative website for indefinite/definite integration, derivatives, plots, solving equations, matricies, etc.
 
  • #15
dextercioby
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I agree quickmath can perform complicated definite integrals, while Mathematica's web integrator can't.

Daniel.
 
  • #16
arildno
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Yes, just multiply both the denominator and the numerator by #sinh t# and then use the ch^2 -sh^2 =1 identity.

Daniel.

Alternatively, use a u=tanh(t/2) substitution for the fun of it.
 

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