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Integral (cos x)^2 dx answer

  1. Jul 23, 2014 #1
    1. The problem statement, all variables and given/known data
    ∫cos2x dx

    3. The attempt at a solution

    I know the answer, and i know how to get there using:

    cos2x+sin2x=1
    cos2x-sin2x=cos2x
    cos2x=(1+cos2x)/2

    But why cant i use the chain rule? Can i?
     
  2. jcsd
  3. Jul 23, 2014 #2

    HallsofIvy

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    I am puzzled by your question. The chain rule is a method for differentiation. Why would you expect to be able to use it to integrate?

    Integration is the opposite of differentiation. Perhaps you are asking "why can't I reverse the chain rule to integrate this?".

    The chain rule says that if x is a function of some variable, t, and y is a function of x, then [tex]dy/dt= (dy/dx)(dx/dt)[/tex]. To use it we have to multiply by dx/dt. To do that in reverse, which is "substitution", the "dx/dt" term has to already be in the integral, you can't just insert it.
     
    Last edited: Jul 23, 2014
  4. Jul 23, 2014 #3
    u=cosx

    therefore...

    ∫(cos2x)dx = ∫(u2)dx = (1/3)u3/u' + C = (1/3)cosx3/(cosx)' + C = (1/3)cosx3/(-sinx) + C

    chain rule reversed?
     
  5. Jul 23, 2014 #4

    HallsofIvy

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    Your second equality, [itex]\int u^2 dx= (1/3)u^3/u'+ C[/itex] is incorrect for precisely the reasons I explained in my first response.
     
  6. Jul 24, 2014 #5
    You must also consider the variable of integration. Since you've made a substitution [itex] u [/itex], you want to integrate with respect to [itex] u [/itex], and as HallsofIvy said, this is moot because of the chain rule.

    Ever heard of integration by parts? You should rather think of this as a product, [itex]\int\cos{(x)}\cos{(x)}dx[/itex]. Try integrating this

    [itex] (fg)' = f'g+ g'f [/itex]
     
    Last edited: Jul 24, 2014
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