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A SIMPLE integral, but I want a nice way to solve.

  1. Aug 24, 2011 #1
    1. The problem statement, all variables and given/known data

    p(t) = cos(t) + (1)/(cos^2(t)

    Find antiderivative.

    2. Relevant equations



    3. The attempt at a solution

    Alright, I'm just starting calc 2. I thoroughly covered a lot of integration involving u substitution, and we have not started that in calc 2 yet. U substitution is so ingrained into my mind right now.

    Wolfram gives sin(t) + tan(t) which I agree with by quickly going through differentiation, but what rules of integration would you apply? I'm guessing some algebra or trig identity?
     
  2. jcsd
  3. Aug 24, 2011 #2
    That p(t) is exactly equal to cos(x) + sec2(x). This can be done simply with direct integration.
     
  4. Aug 25, 2011 #3
    You can also solve it like this:

    [tex]p(t) = cos(t) + \frac 1{cos^2(t)}[/tex]

    Substitute u=cos(t):

    [tex]p(t)=u+u^{-2}[/tex][tex] \int p(t)= \int u+u^{-2}=...[/tex]

    That's a possible way if you wan't to use substitution. :smile:
     
  5. Aug 25, 2011 #4

    SammyS

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    If u=cos(t) , then du = -sin(t) dt . Furthermore, [itex]\displaystyle \sin(t) = \sqrt{1-\cos^2(t)} = \sqrt{1-u^2}\,.[/itex]

    Therefore, [itex]\displaystyle \int p(t)\,dt= -\int \left(u+u^{-2}\right)\sqrt{1-u^2}\,du\,.[/itex]

    So, that's probably not the way to go.

    See Yuqing's post for a straight forward method.
     
    Last edited: Aug 25, 2011
  6. Aug 25, 2011 #5

    HallsofIvy

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    Sorry but that's invalid specifically because that is a "substitution" but you ignored the "dx" and "du". If [itex]u= cos(t)[/itex] then [itex]du= -sin(t)dt[/itex]. To integrate [itex]\int (cos(t)+ 1/cos(t)) dt[/itex] and letting u= cos(t) you would have [itex]\int (u+ 1/u^2)dx[/itex] and substituting for dx will give you a very complex integral.
     
  7. Aug 25, 2011 #6

    gb7nash

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    Here's a solution. However, it does require the circular reasoning that the derivative of tan is sec2. If you can ignore that though, it's legit:

    [tex]\int sec^{2}x dx = \int \sec^{2}x \frac{\tan x}{\tan x} dx[/tex]

    Letting u = tanx, du = sec2x dx, we get:

    [itex]\int \frac{u}{u} du = \int du = [/itex] ...
     
  8. Aug 26, 2011 #7

    HallsofIvy

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    But the "tan x/tan x" is irrelevant. All you are doing is saying [itex]d(tan(x)= sec^2(x)dx[/itex] so [itex]\int sec^2(x)dx= \int d(tan x)= tan(x)+ C[/itex].
     
  9. Aug 26, 2011 #8

    SammyS

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    I had a student who would routinely carry out his substitutions this far. At first it caught me off guard. Then I caught on to what he was doing. He would pick u so that du/dx was equal to the whole integrand, so that he would get: [itex]\displaystyle \int\,\frac{du}{dx}\,dx=\int\,du=u+C\,.[/itex]

    From that time on, I would often refer to this as "Huang's method" , crediting that student.
     
    Last edited: Aug 26, 2011
  10. Aug 26, 2011 #9

    gb7nash

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    Good point. I didn't realize this.
     
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