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Integrals involving Secant & Tangent Derivation

  1. May 13, 2012 #1
    1. The problem statement, all variables and given/known data
    If the power of the secand is even and positive..
    [tex]\int sec^{2k} x tan^{n} x dx = \int (sec^2 x)^{k-1} tan ^n x sec^2 x dx [/tex]


    3. The attempt at a solution

    The way I see it,

    [tex] sec^{2k} x = sec^2 x dx * sec^k x dx[/tex]

    the next step seems to be to break down sec^k, but on closer introspection, the break down goes something like this

    [tex]sec^k x dx = (sec^2 x dx)^{k-1} = sec^{2k-2}x dx[/tex]

    I'm having trouble intuitively accepting that something like 2k-2 = k. It doesn't seem to add up, there seems to be something a bit more complex into it that is left out of detail.
     
    Last edited: May 13, 2012
  2. jcsd
  3. May 13, 2012 #2
    You may want to review your exponent rules.
     
  4. May 13, 2012 #3
    [tex] (x^n)^m = x^{nm}[/tex]
    [tex] sec^m x = (sec x)^m[/tex]
    [tex] (sec^2 x)^{k-1} = (sec x)^ {(2)(k-1)}[/tex]
     
  5. May 13, 2012 #4
    OK. If ## x^{nm}=(x^n)^m ## (which is correct), then what is ## x^n\cdot x^m ##?
     
  6. May 13, 2012 #5
    This is right, note that there are two terms on the right integral.

    Umm, I don't think that's right remember that [tex]a^{b}*a^{c}=a^{b+c}[/tex]

    I also don't know where your [tex]dx[/tex] terms are coming from there.

    You seem to have multiplied your exponents correctly, but forgot that there is another [tex]sec^{2}x[/tex] term in the right integral up top.

    When doing this integral

    [tex]\int sec^{2k} x tan^{n} x dx = \int (sec^2 x)^{k-1} tan ^n x sec^2 x dx [/tex]

    From here there is a handy trig identity involving [tex]sec^{2}x[/tex] that you can use to make it into something easy to use u substitution on.
     
  7. May 13, 2012 #6
    [tex] = x^{n+m}[/tex]
     
  8. May 13, 2012 #7
    I tend to do silly mistakes sometimes, just ignore them.

    Oh there you go, now I get back the original expression. Thanks. ^.^

    I know about u substitution, the calculation is fairly trivial to me. What I was primarily concerned with is the derivation.
     
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