Is There a Trick to Integrating with Law of Cos?

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SUMMARY

The discussion focuses on the challenges of integrating the expression ∫∫∫ r² / (r² + b² - 2rb cosθ) dr sinθ dθ dΦ to find the electric field or potential in electrostatics. The user expresses difficulty with standard integration techniques such as substitution and integration by parts. They suggest that applying the law of cosines, where cosθ equals adjacent/hypotenuse (b/r), could simplify the denominator to r² - b². This insight is crucial for simplifying the integral and improving integration outcomes.

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Students and professionals in physics, particularly those dealing with electrostatics, as well as anyone seeking to improve their integration skills in complex scenarios.

cbirk
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I hope I am posting this in the correct area. This is not specificly a homework question, but something that keeps stumping me on numerous Electrostatic problems.

When attempting to Intigrate for finding the electric field or potential, I frequently end up with an integral over the form ∫∫∫ [itex]r^{2}[/itex] / ( [itex]r^{2}[/itex] + [itex]b^{2}[/itex] - 2rb cosθ) dr sinθ dθ dΦ

I was wondering if there is some sort of method the integrate this that I seem to be missing. A basic U sub doesn't work, and I can't get Integration by parts to do anything but make it worse. I have never been very good at remembering the trig tricks to integration so I was really hoping someone could help me out here. I am lost, and this is costing me lots of points on homeworks and tests.

Thanks in advance.
 
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Also after being stumped for days, I am pretty sure that the way the law of cosines is layed out... that cosθ is equivalent to Adjasent / Hyp = b/r which would simplify the entire denominator to a simple [itex]r^{2}[/itex] - [itex]b^{2}[/itex]

This look at all correct? I have added a picture so you can see the layout

stuff.jpg
 

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