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Trigonmetric Integration Question

  1. Nov 8, 2006 #1
    Hi, I'm new to this forum. I figure it would be appropriate to briefly introduce myself before asking my question.

    My name is Nick. I've graduated highschool and I am presently taking the year off before attending university next fall. Having a lot of free time has allowed me to begin learning Calculus II topics that I'll inevitably encounter in the future.

    Anyways, on to my question(s):

    [tex]\int {{-1}\over{cos^2(x)}} dx[/tex]
    [tex]= -\int {{sin^2(x) + cos^2(x)}\over{cos^2(x)}} dx [/tex]
    [tex]= -\int {{sin^2(x)}\over{cos^2(x)}} dx -\int {{cos^2(x)}\over{cos^2(x)}}dx [/tex]
    [tex]= -\int tan^2(x) dx - \int dx [/tex]
    [tex]= ?[/tex]

    Alas, I do not know how to properly integrate [tex]tan^2(x)[/tex]

    I am relatively confident, however, that my above reasoning is corect. This is because the answer to the following integral is supposed to be:

    [tex]\int tan^2(x)dx = tan(x) - x + C[/tex]


    [tex]\int {{-1}\over{cos^2(x)}} dx[/tex]
    [tex]= -\int tan^2(x) dx - \int dx [/tex]
    [tex]= -tan(x) + x - C - x + D [/tex]
    [tex]= -tan(x) + K [/tex]

    Which should be the correct answer. Thus, my problem is in integrating the tangent x squared function.

    Thank-you in advance,


    Edit: It appears this forum uses a different tex protocol than what I am used to on another forum I frequent. I'm attempting to figure this out, but I do appologize for the confusing mess above.

    Edit 2: I believe I found the culprit. The textit code does not apply on this forum. :)
    Last edited: Nov 8, 2006
  2. jcsd
  3. Nov 8, 2006 #2
    [tex] \int \tan^{2}x dx= \int \sec^{2}x-1\ dx = \tan x - x + C [/tex]
    Last edited: Nov 8, 2006
  4. Nov 8, 2006 #3
    Ugh... now don't I feel rather stupid.

    Trigonmetric identities will be the death of me.
  5. Nov 8, 2006 #4


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    Yeah but [tex]\int \sec^2x dx[/tex] is precisely what he's trying to integrate. The question is, how can it be seen that tanx is a primitive of sec²x?
  6. Nov 8, 2006 #5


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    I guess you're just assumed to have differentiated tanx before and you remembered that is gave sec²x.
  7. Nov 9, 2006 #6

    I knew [tex]{d\over{dx}} tanx = sec^2x [/tex]. :)
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