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Antiderivative calculator?

  1. Jan 12, 2005 #1
    Does anyone know of a good, preferably flash or html-based antiderivative calculator? I'm on a Mac and can't run any windows executable file.

    And if not, would someone please tell me the antiderivative of tan?

    [edit]
    I'll just give the problem I'm working on. I'm trying to find the integral.

    [tex]
    \int_{0}^{\pi/4}\frac{1 + \cos^2\theta}{\cos^2\theta}d\theta
    [/tex]

    Subbing in [tex]\sin^2\theta[/tex] for [tex]1 + \cos^2\theta[/tex], I get [tex]\int_{0}^{\pi/4}\frac{\sin^2\theta}{\cos^2\theta}d\theta[/tex] , which I simplified to [tex]\int_{0}^{\pi/4}\tan\theta d\theta[/tex]

    Assuming I'm correct up to this point, all I need is the antiderivative of tangent to complete the problem.
     
    Last edited: Jan 12, 2005
  2. jcsd
  3. Jan 12, 2005 #2

    dextercioby

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    Trust me,the best antiderivative calculator will always be the human mind.I've heard that Wolfram's "Mathematica" can make wonders... :tongue2: But of course,it's still human made...

    Apply the definition of tangent.Pay attention with the domains of the functions.

    Daniel.
     
  4. Jan 12, 2005 #3
    Edited the original post with the problem I'm working on. I'm sorry, but it's late and I've been doing this pretty much all day. By definition of tangent, do you mean [tex]\sin\theta/\cos\theta[/tex]?
     
  5. Jan 12, 2005 #4

    dextercioby

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    Yes,i meant that definition.

    Wow,there are a lot of mistakes in what u did up there...First of all,u need to understand that
    [tex] \sin^{2}\theta\neq 1+\cos^{2}\theta [/tex] (1)

    Split you integral into two simpler ones...


    Daniel.
     
  6. Jan 12, 2005 #5
    Sorry, I was thinking of [tex]1-\cos^{2}\theta[/tex]. Totally went the wrong way about it.

    [tex]
    \int_{0}^{\pi/4}\sec^2\theta + 1
    [/tex]

    Antiderivative of which is [tex]\tan\theta + \theta[/tex], and the answer is 1 + [tex]\pi/4[/tex]. Major brain hiccup here. Thanks again!

    Antiderivative calculator?
     
    Last edited by a moderator: Feb 17, 2008
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