Integrating Trigonometric Functions with Evasive Substitutions

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Homework Help Overview

The discussion revolves around integrating a trigonometric function, specifically the integral of the form \(\int \frac{\sin(at)}{(1 + b\sin^{2}(at))^{1/2}} dt\). The original poster is exploring this integral in the context of modeling a particle in an oscillating field and is seeking guidance on potential substitutions or methods to approach the problem.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the challenges posed by the sine function in the integral and consider various substitutions. One participant suggests transforming the expression \(1 + b\sin^2(at)\) into a form involving cosine, proposing a substitution involving \(u = \cos(at)\) to simplify the integral.

Discussion Status

The discussion is active, with participants sharing insights and suggestions for substitutions. There is acknowledgment of the complexity of the integral, and while no consensus has been reached, a potential direction involving a trigonometric substitution has been proposed.

Contextual Notes

The original poster mentions feeling rusty with integral solving techniques and is looking for hints or resources to improve their understanding of similar integrals.

Rory9
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I am staring at an integral of the form

[tex] \int \frac{sin(at)}{(1 + bsin^{2}(at))^{1/2}} dt[/tex]

which I have generated for myself (in attempting to model the behaviour of a particle in an oscillating field). I can't see a sensible substitution to try, at present. I could hunt down a standard integral, perhaps, but I suspect something obvious is evading me...

Any hints? Also, any suggestions for brushing up on solving integrals of this sort? I'm a bit rusty :-)

Cheers!
 
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Rory9 said:
I can't see a sensible substitution to try, at present.
It's sin that makes this complicated, isn't it? That seems the obvious place to start.
 
This might be useful: Change 1 + b*sin^2(at) into 1 + b*(1 - cos^2(at) = 1 + b - b*cos^2(at).

You could then use the substitution u = cos(at), du = -a*sin(at)dt.

Then your integral would be roughly du/(A - bu^2)^(1/2), and you might be able to find that in a table of integrals or, failing that, apply a trig substitution.

Anyway, that's the direction I would go as a start.
 
Mark44 said:
This might be useful: Change 1 + b*sin^2(at) into 1 + b*(1 - cos^2(at) = 1 + b - b*cos^2(at).

You could then use the substitution u = cos(at), du = -a*sin(at)dt.

Then your integral would be roughly du/(A - bu^2)^(1/2), and you might be able to find that in a table of integrals or, failing that, apply a trig substitution.

Anyway, that's the direction I would go as a start.

A good suggestion. Thank you. It will end up with an arc sin of a cos, I think, but perhaps that can be rewritten more elegantly...
 

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