Integration by Parts

  • Thread starter hex.halo
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  • #1
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Homework Statement



Use integration by parts to find:

y=... if dy=arcsinh(x) dx

Homework Equations



int(v.du)=uv-int(u.dv)

The Attempt at a Solution



I understand how to perform integration by parts. My problem here is, what are my 'v' and 'du'?
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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[tex]\int sinh^{-1}x dx[/tex]


well [itex]dv\neq sinh^{-1}x dx[/itex] since to find v you'd need to integrate that and well obviously you can't do that. So [itex]u=sinh^{-1}[/itex] and [itex]dv=1dx[/itex].
 
  • #3
A good general tip is to take very hyperbolic function and every trig function and learn how to differentiate it or integrate it. That way you don't have to remember the tables of functions. If you like memorising tables though do it that way, but do both. :smile: Just a really good but obvious hint I picked up recently that I thought might be useful.

It's generally a good idea to try and split arc functions into there [itex]\frac{1}{\text{trig function}}[/tex] equivalents first, I don't know if it's just me but that seems to work out better more often than not? Clearly here the substitution becomes much easier when you do this, but I find it's a good general rule..?
 
Last edited:
  • #4
tiny-tim
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I understand how to perform integration by parts. My problem here is, what are my 'v' and 'du'?

Hi hex.halo ! :smile:

You have to do a substitution first, and then integrate by parts!

Put x = sinha, dx = cosha da …

and you get … ? :smile:
 

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