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    Gernald's Avatar
    Gernald Posts: 901, Reputation: 93
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    #1

    Mar 26, 2009, 10:34 AM
    antiderivitive c and d
    Hi all!
    I'm trying to find the anti derivative of a(t)=cos(t)+sin(t)

    so far I have sin(t)+c-cos(t) as the antiderivitive but Im confused because the antiderivitive of cos(t) is sin(t)+c and the antiderivitive of sin(t) is -cos(t)+c
    so should the antiderivitive of a(t) be sin(t)+c-cos(t)+D??
    just confused about the c/d thing.

    Thanks!
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #2

    Mar 26, 2009, 11:00 AM
    The antiderivative of would be



    No need for another constant.



    Now, when we take the derivative of sin(t)-cos(t)+C we get back where we started. To cos(t)+sin(t).

    Because the derivative of a constant, C, is 0.
    ngasnier's Avatar
    ngasnier Posts: 30, Reputation: 4
    Junior Member
     
    #3

    Mar 27, 2009, 04:31 PM
    Quote Originally Posted by Gernald View Post
    Hi all!
    I'm trying to find the anti derivitive of a(t)=cos(t)+sin(t)

    so far I have sin(t)+c-cos(t) as the antiderivitive but Im confused because the antiderivitive of cos(t) is sin(t)+c and the antiderivitive of sin(t) is -cos(t)+c
    so should the antiderivitive of a(t) be sin(t)+c-cos(t)+D ???
    just confused about the c/d thing.

    Thanks!

    It is impossible to identify the constant of integration. The solution which you seek lacks a closed form .
    Gernald's Avatar
    Gernald Posts: 901, Reputation: 93
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    #4

    Mar 28, 2009, 11:29 AM
    Quote Originally Posted by ngasnier View Post
    It is impossible to identify the constant of integration. The solution which you seek lacks a closed form .
    Huh??
    I'm just going to go with +c and be done with it.

    Thanks anyway.

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