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Somewhat easy integral

  1. Apr 7, 2008 #1
    So this is my first year of calculus and we're doing integrals using u-substitution

    I'm having trouble with this integral:

    integral of (1/x^2)*sec(1/x)*tan(1/x)dx

    i let u=(1/x)
    so du=(1/x^2)dx
    so du(x^2)=dx

    If I do this, the integral works out perfectly since the x^2 cancel out

    so it equals sec(u) + C

    HOWEVER, my question is
    if i let u = (1/x)

    why don't i replace the (1/x^2) (in the beginning of the integral) with u^2

  2. jcsd
  3. Apr 7, 2008 #2
    First of all, the derivative of [tex]\frac{1}{x}[/tex] is not [tex]\frac{1}{x^2}[/tex]

    Second of all, if [tex]u^2 = \frac{1}{x^2}[/tex]

    then [tex]2udu = \frac{-2}{x^3}dx[/tex]

    And you haven't made anything simpler for yourself.
  4. Apr 7, 2008 #3
    the derivative of 1/x is -1/x^2 correct? sorrry

    but if i were to do a u substitution like

    integral of tan^4(x)*sec^2(x)

    and i let u=tan x
    then du*sec^2x=dx

    and that leads to the integral of u^4, correct??? i'm confused
  5. Apr 7, 2008 #4
    I don't follow. The derivative of u with respect to x is du / dx and the derivative of tan x is (secx)^2. so du / dx = (secx)^2.
  6. Apr 7, 2008 #5
    right. so dx = du / (secx)^2 so
    if i substitute that back into the original integral the (secx)^2 cancel out.
    is the clearer?
  7. Apr 7, 2008 #6
    But you can't let u = tan(x) because there is no tan(x) in the integral. There's a tan(1/x).
  8. Apr 7, 2008 #7
  9. Apr 7, 2008 #8
    The difference is that you are saying "let u^2 = some function, whereas in this question, we let u = some function, and when you substitute u back into the integral, it just happens to become u^3.

    What you want to do is let u = some function so that du = (some function that's already inside the integral)dx. This simplifies things in terms of u and du with no X's.
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