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Homework Help: Antiderivative quicky

  1. Feb 13, 2007 #1
    1. The problem statement, all variables and given/known data
    whats the integral of cos^2(x)



    2. Relevant equations
    ?



    3. The attempt at a solution

    is it just sin^3(x)/3?
     
  2. jcsd
  3. Feb 13, 2007 #2

    cepheid

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    No. And you know that that is the wrong answer because:

    [tex] \frac{d}{dx}\left(\frac{\sin^3(x)}{3}\right) = \sin^2(x)\frac{d}{dx}(\sin x) = \sin^2(x)\cos(x) \not= \cos^2(x) [/tex]

    Notice how I had a composition of functions and used the chain rule to differentiate. How do you deal with integrating something that has a composition of functions?
     
  4. Feb 13, 2007 #3

    mjsd

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    "easiest" way to integral this would probably be using the double angle formula
     
  5. Feb 13, 2007 #4

    Gib Z

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    Write out cos^2 x as (1-cos 2x)/2 is what mjsd meant.
     
  6. Feb 14, 2007 #5
    integrtion by parts:

    http://www.artofproblemsolving.com/Forum/latexrender/pictures/108554a051d314e3db05f4254cd279b0.gif [Broken]
     
    Last edited by a moderator: May 2, 2017
  7. Feb 15, 2007 #6

    Gib Z

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    That seems abit longer and harder than msjds method.
    His way easily goes to this:
    [tex]\frac{1}{2} \int 1 dx + \int \cos 2x dx = \frac{1}{2} (x + \int cos 2x dx)[/tex]

    u=2x du/dx = 2
    [tex]\frac{1}{2}(x+\frac{1}{2}\int cos u du)= \frac{1}{2} ( x+\frac{1}{2}\sin u)[/tex]


    We're done, pretty fast too :)
    [tex]\int \cos^2 x dx = \frac{x}{2} + \frac{\sin 2x}{4}[/tex]

    BTW emilgouliev, nice work, and Welcome to Physicsforums. :)
     
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