Help with Integral: c*Sin[ArcTan[f(t)]]

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  • Thread starter Thread starter FunkyDwarf
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    Integral
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Discussion Overview

The discussion revolves around the integral of the expression c*Sin[ArcTan[f(t)]] with respect to t, where f(t) is specified as a linear function of t. Participants explore various substitutions and approaches to simplify or solve the integral.

Discussion Character

  • Exploratory, Mathematical reasoning, Homework-related

Main Points Raised

  • One participant expresses difficulty in progressing with the integral after several substitutions, indicating a lack of inspiration.
  • Another participant restates the integral in a more formal mathematical notation.
  • A different participant provides a reminder about the relationship between tangent and sine, suggesting a method to solve for sin(arctan(x)).
  • A later reply acknowledges a concern about applying inverse sine correctly but reassures that it can be done due to known relationships from the substitution.

Areas of Agreement / Disagreement

The discussion does not reach a consensus, as participants are exploring different methods and expressing varying levels of understanding and concern regarding the application of mathematical principles.

Contextual Notes

Participants do not clarify specific assumptions about the function f(t) beyond it being linear, and there are unresolved steps in the mathematical reasoning presented.

FunkyDwarf
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Hey guys,

im completely stumped on this one. I had to go through several substitutions just to get it to this stage which is uninspiriring :P

Integral wrt t of c*Sin[ArcTan[(f(t)]] where f(t) is a linear function of t

Sorry, i can't use latex to save my life :p
-G
 
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[tex]\int c*\sin{(\arctan{f(t))}} dt[/tex]?
 
Remember:
[tex]x=tan(\arctan(x))=\frac{\sin(\arctan(x))}{\sqrt{1-\sin^{2}(\arctan(x)}}[/tex]

Solve for sin(arctan(x))
 
Ah yeh of course, i was worried about applying inverse sin to the top and bottom equally but you sort of can because you know what sine is from the substitution and so can get cos. Thanks!
 

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