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Trig Integration Question

  1. Feb 12, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int^{\pi}_{0}(cos(x))^{6}dx[/tex]

    2. Relevant equations

    * Half-Angle => (cos(x))^{2} = (1/2)(1 + cos(2x))

    3. The attempt at a solution

    We just started this chapter today and during lecture the only example of this form (even powers/cosine) was (cos(x))^{2}, which only requires integrating the Half-Angle Formula. The way I approached this problem looks like it's taking me towards pretty big mess:

    [tex]\int^{\pi}_{0}((cos(x))^{2})^{3}[/tex] = [tex]\int^{\pi}_{0}[(1/2)(1+cos(2x))]^{3}[/tex]

    Any suggestion?
     
  2. jcsd
  3. Feb 12, 2009 #2
    How about using hyperbolic cosh to get the things into full to e's?
     
  4. Feb 12, 2009 #3

    Dick

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    Homework Helper

    The way to avoid the mess is to use a formula to express the integral of cos^n(x) in terms of an integral of cos^(n-2)(x). See for example http://www.sosmath.com/calculus/integration/powerproduct/powerproduct.html at the bottom of the page. You should also note that you have a DEFINITE integral from 0 to pi. You don't need to evaluate the full indefinite integral. The term involving sin(x) in the formula vanishes in your case. If you don't already know that formula, you should probably try to prove it. It's not too hard. Just integration by parts.
     
  5. Feb 12, 2009 #4

    HallsofIvy

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    Tedious but not that big a mess:
    [tex]=\frac{1}{8} \int_0^\pi 1+ 3cos(2x)+ 3cos^2(2x)+ cos^3(2x) dx[/tex]
    you can immediately integrate [itex]\int 3cos(2x) dx[/itex] and writing [itex]cos^3(2x)[/itex]as [itex]cos^2(2x)cos(x)= (1- sin^2(x))cos(x)[/itex] lets you integrate [math]\int cos^3(2x)dx[/itex].

    The only "hard" part is
    [tex](3/8)\int_0^\pi cos^2(2x)dx[/tex]
    and you can use [tex]cos^2(2x)= (1/2)(1+ cos(4x))[/tex] for that.
     
  6. Feb 12, 2009 #5
    My prof covered this whole chapter today and only introduced two "cases" that can be identified for solving these problems, each case having two methods which are chosen by looking at the even or odd powers.

    case 1: integral[((cosx)^m)*(sinx)^n))dx]

    - or -

    case 2: integral[((tanx)^n)*((secx)^m)dx]

    I figured these were the only ways to solve these problems but I'm open to any new methods. I'm not clear on what you are telling me to do, are you saying to somehow get it in the form:

    (e^x+e^x)/2

    **sorry this reply was to "rootx", it took me awhile to post I didn't see the other replies after
     
  7. Feb 12, 2009 #6
    Dick, thanks for the link, I forgot about that reduction formula.
     
  8. Feb 12, 2009 #7
    I was thinking of:
    (cos (x))^6 = [(exp(i.x) + exp(-i.x))/2]^6
    It's easier to expand the right side and not hard to integrate.
    I remember using this several times somewhere but I have forgotten where:frown:

    But, those reductions formulas are best here.
     
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