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Math Help - integration problem - antiderivative??

  1. #1
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    integration problem - antiderivative??

    how is the integral of sin 2t = -1/2cos 2t?

    i cant seem to get where the 1/2 comes from.
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  2. #2
    Newbie Homeomorphism's Avatar
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    \int{\sin{a}x} = -\dfrac{1}{a}\cos{a}x.
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  3. #3
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    It might help to look at the problem in reverse:

    \frac{d}{dt}\left(-\frac{1}{2}\cos{2t}\right)=-\frac{1}{2}(-sin{2t})\frac{d}{dt}(2t)=sin{2t}

    The 1/2 is there to cancel the 2 coming from the \frac{d}{dt}(2t) part of the chain rule.

    Post again in this thread if you're still having trouble.
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by calculus0 View Post
    how is the integral of sin 2t = -1/2cos 2t?

    i cant seem to get where the 1/2 comes from.
    If you do not understand the direct formula, try substitution method:

    let u = 2t then du = 2 dt \rightarrow dt = \frac{du}{2}

    \int sin2t dt  = \int sin(u)\frac{du}{2} = \frac{1}{2} (-cos(u))+C

    substitute u = 2t in the above answer"

     - \frac{1}{2} cos2t+C
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