Need integral help, i think its trig sub., thanks

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    Integral Thanks Trig
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Discussion Overview

The discussion revolves around solving the integral \(\int_{0}^{a}\frac{x}{(x^2+z^2)^\frac{3}{2}}dx\). Participants explore various substitution methods, including trigonometric substitution and algebraic substitution, to simplify the integral.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant initially presents the integral and expresses uncertainty about the best method to solve it.
  • Another participant suggests the substitution \(x^2 = t\) as a potential approach.
  • A different participant reports success using the trigonometric substitution \(x = z \tan{\theta}\), indicating it worked for them.
  • Another participant challenges the convenience of the trigonometric substitution, suggesting that the algebraic substitution \(x^2 = t\) leads to a simpler form for the numerator.
  • One participant proposes using the substitution \(t = x^2 + z^2\), which leads to a different integral expression, specifically \(\frac{1}{2}\int_{z^2}^{z^2 + a^2} t^{-3/2}dt\).
  • Several participants express that the alternative substitutions are progressively easier to work with.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best substitution method, as multiple approaches are suggested and evaluated. Some participants find certain substitutions easier than others, indicating a variety of preferences.

Contextual Notes

The discussion includes various assumptions about the convenience and effectiveness of different substitution methods, but these assumptions are not universally agreed upon. The effectiveness of each substitution may depend on the specific context of the integral.

klawlor419
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\int_{0}^{a}\frac{x}{(x^2+z^2)^\frac{3}{2}}dx

Hi all, I'm stuck on this one. Sure there's an easier way to do it. Don't have my calc book currently. Thanks!
 
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Put x2 = t.
 
Got it with the trig sub x=z \tan{\theta}
Thanks!
 
klawlor419 said:
Got it with the trig sub x=z \tan{\theta}
Thanks!

No, that substitution isn't convenient,

If you substitute x2 = t, you would get the numerator as dt (since dt= 2xdx)
 
Thanks! That is slightly easier!
 
I would have used t= x^2+ z^2 so that dt= 2x dx, (1/2)dt= xdx.

The integral becomes
\frac{1}{2}\int_{z^2}^{z^2+ a^2} t^{-3/2}dt
 
Thanks! Thats slightly easier still!
 

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