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Simple antiderivative

  1. Mar 23, 2007 #1
    Find f if f '(t)=8cost+sec^2t and f(pi/3)=4 .

    so f(t)=8sint+tant+c



    so f(t)=8sint+tant-3^1/2 is that right
  2. jcsd
  3. Mar 23, 2007 #2
    You need to re-evaluate the values of sin and tan at pi/3.
  4. Mar 23, 2007 #3
    thank you very much
  5. Mar 23, 2007 #4
    sin(pi/3) is sqroot3/2 and tan(pi/3) is sqroot3 and 8(sqroot3/2)+sqroot3 is

    is that right?
  6. Mar 23, 2007 #5

    But for the sake of clarity, use something like sqroot(3)/2 instead of sqroot3/2 when referring to [tex]\frac{\sqrt{3}}{2}[/tex]. :smile:
  7. Mar 23, 2007 #6
    ok. so now i have (5sqroot(3)/2)+c=4

    then c=4-5sqroot(3)/2 so that would be -sqroot3

    is that right
  8. Mar 23, 2007 #7
    Why the 2 in the denominator?

    I don't understand. You've already solved for c, so...

    f(t) = 8sin(t) + tan(t) + 4-5sqroot(3)
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