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Calc integration

  1. Apr 25, 2005 #1
    question and work is here

    I got the answer down to letter A and D. Now I feel like it's letter A but not sure...
     
  2. jcsd
  3. Apr 25, 2005 #2
    Youre asking if [tex]\frac{sin^2(x)}{2} = \frac{-cos(2x)}{4} [/tex]

    Try x = 0

    Or notice that f(x) = sin(x)cos(x) = sin(2x)/2, then the integral is real easy.
     
  4. Apr 25, 2005 #3

    dextercioby

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    The easiest method is to differentiate each of the 3 results...

    "D" is the correct answer.

    Daniel.
     
  5. Apr 25, 2005 #4
    Take the derivative of each of the choices.

    *Haha, too late
     
  6. Apr 25, 2005 #5
    Not at all. If F(x) is an antiderivative of f(x) then so is F(x)+C for any C. Here,

    [tex]\frac{\sin^2 x}{2} = \left(-\frac{\cos 2x}{4}\right) + \frac{1}{4}[/tex]
     
    Last edited: Apr 25, 2005
  7. Apr 25, 2005 #6
    how did you get III to be true?
     
  8. Apr 25, 2005 #7
    look at my last post.
     
  9. Apr 25, 2005 #8
    Convert f(x) to what I recommended and the integral evaluates to D directly.
     
  10. Apr 25, 2005 #9
    Good call, I didnt see that.
     
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