Integration by Parts: Solving y = arcsinh(x)

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Homework Help Overview

The discussion revolves around the application of integration by parts to solve the integral of the function y = arcsinh(x). Participants are exploring the necessary components for the integration process.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning the identification of 'v' and 'du' in the integration by parts formula. There are suggestions to consider substitutions and the differentiation or integration of hyperbolic functions.

Discussion Status

The discussion is active, with participants sharing insights and tips about integration techniques. Some guidance has been offered regarding substitutions, but there is no explicit consensus on the best approach yet.

Contextual Notes

Participants mention the challenge of integrating arcsinh(x) and the need for substitution before applying integration by parts. There is an emphasis on understanding hyperbolic functions and their derivatives.

hex.halo
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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'?
 
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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].
 
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..?[/itex]
 
Last edited:
hex.halo said:
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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