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Integration Problem

  1. Dec 5, 2007 #1
    [tex]\int[/tex](tan x)^2(sec x)^2
    The book shows the integration going straight to
    (tan x)^3/3
    Which I can see how to get that, simply raise the tangent to +1 power & then divide by three, but I'm not seeing where the secant cancels out. I checked the back of the book integrals but couldn't find anything there. Any ideas?
  2. jcsd
  3. Dec 5, 2007 #2


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    Try a substitution u=tan x. What is du/dx?
  4. Dec 5, 2007 #3
    Thanks, its sometimes unbelievable the little things you over look. I even tried to set u=sec x... but for some reason never thought of doing that to tangent.
  5. Dec 5, 2007 #4

    Gib Z

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    The fact that you have forgotten to have the differential (dx) in your integral shows you aren't too familiar with integration by substitution =[ Try revising your textbook!
  6. Dec 5, 2007 #5
    What are you talking about? If u=tan x, then du=(sec x)^2, which cancels the secant in the equation perfectly.

    Then to integrate (tan x)^2, simply add one to the power, (tan x)^3 & then divide by three.
  7. Dec 5, 2007 #6

    Gib Z

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    I know, but in your first post you didn't write the dx in your integral, which is crucial to actually 'cancel' with the dx in the du/dx term.
  8. Dec 5, 2007 #7
    You're right, I should have written the dx, but its more of a minor annoyance, so long as you're canceling them out in your head. I read your post incorrectly, when you said "to have," I read "to halve." The correct English to be "forgotten the differential." :)
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