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General help for trigonemtric u-substitution

  1. Apr 13, 2012 #1
    I'm currently in calculus 2 and we are learning about u-substitution and I am having a little bit of difficulty seeing how to start problems that involve trig functions. I know sometimes I can rewrite them to help suit my needs but this isn't always evident. I have for example the indefinite integral of sec^3(X)tan(x) and am pretty lost where to start something like this. Would I need to rewrite these terms or I can I just use u=sec(x)? If anyone knows of a way to go about these problems in general I would be greatly appreciative.

    Thanks in advance!
     
  2. jcsd
  3. Apr 13, 2012 #2
    The derivative of u and u itself have to get rid of all other terms with x in it, or you'll have an integral in terms of u and x. In this situation, you have a few possibilities. If you are not familiar with the derivatives to see the answer instantly, just iterate through and find it:
    u = x
    u = sec(x)
    u = sec^2(x)
    u = sec^3(x)
    u = tan(x)
    u = tan(x)sec(x)
    u = tan(x)sec^2(x)
    u = tan(x)sec^3(x)

    Just start with the ones that are straight looks into a table of derivatives like tan(x), sec^2(x), etc.

    Remember, substitution is algebraic substitution, even for the dx and du stuff. I think pointing this out demystifies it for students having trouble. So if you have
    [tex] \int f(x)dx = \int g(x)h(x)dx[/tex]
    and you can make a substitution that writes g(x) into G(u) such that
    [tex] \frac{du}{dx} = h(x) \rightarrow du = h(x)dx[/tex]
    then
    [tex] \int G(u) du[/tex]
    So you actually replace "h(x)dx" with "du"

    I mean, this is just the definition, but I want to really stress you are simply saying du equals that stuff, so you replace it with du. We are treating du and dx like variables (multiplying both sides by dx to find what du equals).
     
    Last edited: Apr 13, 2012
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