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How to integrate int F sin^2(At) dt

  1. Sep 5, 2006 #1

    danago

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    Gold Member

    Hey. Ive barely started learning about integration, but during an experiment we conducted for physics, the following came up:

    [tex]
    \int {F\sin ^2 At} dt
    [/tex]

    How would i integrate that? I do understand that it would generally require that i show my working on the issue, but in this case, its a bit hard, since i dont have any working.

    If someone could briefly explain how i would go about evaluating this indefinite integral, i would be greatly appreciative.

    Thanks,
    Dan.
     
  2. jcsd
  3. Sep 5, 2006 #2

    TD

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    Homework Helper

    I assume F and A are constants, so that shouldn't be of any trouble.
    The annoying part is the square of the sine, but you could use:

    [tex]
    \cos \left( {2x} \right) = \cos ^2 x - \sin ^2 x = 1 - 2\sin ^2 x \Rightarrow \sin ^2 x = \frac{{1 - \cos \left( {2x} \right)}}{2}
    [/tex]

    Does that help?
     
  4. Sep 5, 2006 #3

    danago

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    Hmmm...ok i understand that. How would i integrate something like this:

    [tex]
    \int {\cos At} {\rm dt}
    [/tex]

    Would it just be [tex]-sin At+c[/tex]
    ?
     
  5. Sep 5, 2006 #4

    TD

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    Almost, you have to watch out with the A. Hint: dt = 1/A d(At).
    Or, use an explicit substitution: let y = At so dy = Adt so 1/A dy = dt.
     
  6. Sep 5, 2006 #5

    J77

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    ...and you may want to rethink that minus sign.
     
  7. Sep 5, 2006 #6

    danago

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    Gold Member

    ahh so would this be right? :

    [tex]
    \int {\cos At} {\rm dt = }\frac{{\sin At}}{A} + c
    [/tex]
     
  8. Sep 5, 2006 #7

    J77

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    Yep - check by differentiating.
     
  9. Sep 5, 2006 #8

    danago

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    Gold Member

    Awesome :D Thanks

    OK so back to the original question. Is this working right?:

    [tex]
    \displaylines{
    \int {F\sin ^2 At} {\rm }dt = F\int {\sin ^2 At} {\rm }dt \cr
    = F\int {\frac{{1 - \cos (2At)}}{2}} {\rm }dt \cr
    = \frac{F}{2}\int {1 - \cos (2At)} {\rm }dt \cr
    = \frac{F}{2}\left( {\int {1{\rm }dt - \int {\cos (2At)} {\rm }dt} } \right) \cr
    = \frac{F}{2}\left( {t - \frac{{\sin 2At}}{{2A}}} \right) \cr
    = \frac{{Ft}}{2} - \frac{{F\sin 2At}}{{4A}} \cr}
    [/tex]
     
  10. Sep 5, 2006 #9

    J77

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    yep - plus constant, of course.
     
  11. Sep 5, 2006 #10

    danago

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    Gold Member

    Oops forgot about that.

    Anyway, thanks very much for the help. Greatly appreciated :D
     
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