[tex]\int \frac{x}{\left(x^2+z^2\right)^{3/2}} \, dx[/tex]

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SUMMARY

The integral \(\int \frac{x}{\left(x^2+z^2\right)^{3/2}} \, dx\) can be effectively solved using trigonometric substitution, specifically \(x = z \tan \theta\). The discussion highlights the challenges faced by a student in an electrodynamics course when attempting various u-substitutions without success. The use of Mathematica was mentioned as a tool for solving the integral, but the key takeaway is the importance of recalling trigonometric substitution techniques from previous calculus courses.

PREREQUISITES
  • Understanding of integral calculus, specifically techniques of integration.
  • Familiarity with trigonometric identities and substitutions.
  • Experience with software tools like Mathematica for solving integrals.
  • Basic knowledge of electrodynamics concepts related to integrals.
NEXT STEPS
  • Review trigonometric substitution methods in integral calculus.
  • Practice solving integrals using Mathematica for verification.
  • Study the relationship between integrals and electrodynamics applications.
  • Explore advanced integration techniques, including integration by parts and partial fractions.
USEFUL FOR

Students in calculus or electrodynamics courses, educators teaching integration techniques, and anyone looking to enhance their problem-solving skills in mathematical physics.

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Homework Statement


[tex]\int \frac{1}{\left(x^2+z^2\right)^{3/2}} \, dx[/tex]

I have been trying various u-substitutions for about 2 hours now, but I cannot seem to find a way to solve this by hand! I used mathematica to solve the problem. I feel like it will be fairly straightforward once I figure out what u should be, if u-sub is the way to go.

This is for my electrodynamics course, and getting hung up on integrals is not helpful.

I've tried:
u=x^2
u=1/(x2+z2)1/2
u=1/(x2+z2)
u=1/(x2+z2)3/2
u=x/(x2+z2)1/2
u=x/(x2+z2)
u=(x2+z2)1/2
u=(x2+z2)
u=(x2+z2)3/2
 
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Have you tried a trig substitution...something like [itex]x=z\tan\theta[/itex] looks appropriate:wink:
 
wow...never even crossed my mind to use trig substitution. In fact, I completely forgot that as a method. And its so useful! Thank you! I guess I ought to review calc 2 stuff from HS...

Thanks again!
 

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