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Trigonometric Integration and U Substitution

  1. Sep 17, 2008 #1

    We were going over trigonometric integration in Calculus II the other day. I got the basic idea, but get lost when we're doing the u-substitution.

    We had a problem like this:

    [tex]\int cos^3 (x) dx[/tex]

    Then we did:

    [tex]\int (1 - sin^2 (x)) cos(x) dx[/tex]

    Starting u-substitution:

    [tex]u = sin(x)[/tex]
    [tex]du = cos(x) dx[/tex]

    So then we have:
    [tex]\int (1 - u^2) du[/tex]

    All reasonable so far. But then, du just kind of disappears.

    [tex]u - u^3/3[/tex]
    Which is
    [tex]sin(x) - 1/3 sin^3 (x) + c[/tex]

    I see how u-substitution works when its something simple, but I can't see how the answer here can be derived to get what was given before.
  2. jcsd
  3. Sep 17, 2008 #2
    What do you mean du disappears? Do you think dx "disappears" as well when you integrate f(x) dx? I'm not sure you really understand what dx is supposed to represent, in your own words tell me what it is.

    Your other question i'm guessing is how is the derivative of

    [tex] sin(x) - \frac{1}{3} sin^{3}(x) + c [/tex]

    equal to


    Well differentiate your result you get:

    [tex] cos(x) - sin^{2}(x)cos{x} = cos(x) \left(1 - sin^{2}(x) \right) = cos(x) \cdot cos^{2}(x) = cos^{3}(x) [/tex]

    Does that help?
  4. Sep 17, 2008 #3
    I know what dx is, but I didn't really express my question in the best way. If you like though, in terms of integrals, dx is the infinitesimal parts that the area of a curve is split up into.

    The thing is, du isn't just equal to dx. I mean it represents something else, in this case cos(x). Doesn't that matter?

    Also your explanation confuses me. How and why did you separate a cos from the sin^3?

    The most I could figure out is you could use the chain rule:

    [tex]sin(x) - sin^3 (x)[/tex]
    [tex]cos(x) - 3 sin^2 (x) cos(x)[/tex]
    [tex]cos (x) \left(1 - 3 sin^2 (x) \right)[/tex]

    That 3 gets in the way though.

    Basically what I'm looking for is, if you have something expressed in terms of u and du, with no other variables, then it is valid to just discard everything represented by du?
  5. Sep 17, 2008 #4
    You forgot the 1/3 before [itex]sin^3(x)[/itex] thus when you differentiate this term, the 1/3 cancels out and you are left with [itex]sin^2(x)[/itex]
  6. Sep 17, 2008 #5
    You are right du isn't dx, it does represent something else but not cos(x), it reprents cos(x) dx i.e. once again some infit. change but that's because u isn't x, it's now sin(x) so when you were integrating sin^3(x) dx you were looking at the height*width scenario and were letting your width be infit. small so you would take smaller and smaller intervals but letting the number of them go to infinity, now you are changing what you are integrating and so now have to change with respect to what, i.e. you are expressing your height and therefore your width in a different manner. Does this help you understand it?
    You forgot 1/3.
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