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Math help

  1. Sep 11, 2004 #1
    the problem is shown in the picture i attached.

    wouldnt i use [tex]u = 2sec(\theta)[/tex] as a sub?
     

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  3. Sep 11, 2004 #2

    HallsofIvy

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    Well, not "u= 2 sec θ" because there was no "u" in the integral!

    But setting 2x= secθ will work very nicely (Notice where the 2 is!)

    sin2θ+ cos2θ= 1 so, dividing both sides by cos2θ, tan2θ+ 1= sec2θ and then
    sec2θ- 1= tan2θ.

    That's the whole point of the substitution. If 2x= sec θ then 4x2 =
    sec2 θ so 4x2- 1= tan2θ and you can drop the square root: [itex]\sqrt{4x^2-1}= \sqrt{tan^2\theta}= tan \theta [/itex].

    (You might have been thinking of examples involving "x2- 4" where the substitution x= 2sec θ would give 4sec2- 4= 4(tan2 θ)
     
    Last edited: Sep 11, 2004
  4. Sep 11, 2004 #3
    ok so the problem should look like [tex]\int \frac{x^3}{tan \theta}[/tex] right now.

    i dont know what to do next. i cant use u-du. can i use parts on this?

    if you set [tex] 2x= sec \theta[/tex]

    then [tex] x= \frac{sec \theta}{2}[/tex]
    then would i have to take the derv. of that?
    [tex]dx= 1/2*ln(sec(\theta)+tan(\theta)) [/tex]


    then we would rewrite the problem as....(plugged in [tex]x=\frac{sec \theta}{2}[/tex] for x)

    [tex]\int\frac{(\frac{sec \theta}{2})^3}{\sqrt{4(\frac{sec \theta}{2})^2-1}} * 1/2*ln(sec(\theta)+tan(\theta))[/tex]

    hmm my way seems longer and harder...


    can you help me with your way?
     
    Last edited: Sep 11, 2004
  5. Sep 11, 2004 #4

    Fredrik

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    I'm not familiar with the notation "sec", but from what HallsofIvy wrote, I take it that sec θ is just 1/cos θ.

    When I use this, and calculate dx=dx/dθ dθ, the integral simplifies to

    [tex]\frac{1}{16}\int\frac{1}{cos^4\theta}d\theta[/tex]

    I don't know how to solve that, but maybe you do.

    The thing that you're doing wrong is the derivative. There shouldn't be any logarithms in there.
     
  6. Sep 11, 2004 #5

    Pyrrhus

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    Let's review what you got

    [tex] 2x = \sec\theta [/tex]
    [tex] \sqrt{4x^2-1}= tan\theta [/tex]
    [tex] dx = \frac{sec\theta tan\theta d\theta}{2} [/tex]

    so you will have:

    [tex] \int \frac{(\frac{sec\theta}{2})^3}{tan\theta}\frac{sec\theta tan\theta d\theta}{2} [/tex]

    and working the terms:

    [tex] \frac{1}{16} \int sec^4\theta d\theta [/tex]

    [tex] \frac{1}{16} \int sec^2\theta (1+tan^2\theta) d\theta [/tex]

    [tex] \frac{1}{16} \int sec^2\theta d\theta + sec^2\theta tan^2\theta d\theta [/tex]

    [tex] \frac{1}{16} [\int sec^2\theta d\theta + \int sec^2\theta tan^2\theta d\theta] [/tex]

    and you can finish it yourself, i believe...
     
    Last edited: Sep 11, 2004
  7. Sep 11, 2004 #6
    use the fact that 1 + tan² = 1/cos² and write the integrandum as
    (1/cos²(x) * 1/cos²(x))dx.

    Then you also know that d(tan(x)) = dx/cos²(x)

    So you get as integrandum 1/cos²(x)*d(tan(x)) and substitute the cos² by the above expression.

    regards
    marlon (sorry for the writing, i really should start using LaTex)
     
  8. Sep 11, 2004 #7


    [tex] \frac{1}{16} [\int sec^2\theta d\theta + \frac{1}{16}\int sec^2\theta tan^2\theta d\theta] [/tex]

    my work:
    found the anti-derv.



    [tex] \frac{1}{16}(tan \theta) + \frac{1}{16}(sec \theta)^2 [/tex]


    ok now this is the part im confused on... i know i need to draw a triangle and do something with it...
    triangle is attached, but nothing is filled in. i took notes, but i dont know what to fill the values for the triangle.
    im thinking that the values are....
    a(Adjacent) = x^3
    h = 1/2
    O = [tex]\sqrt{(4x^2-1)}[/tex]

    how do you know what the a,h, and o sides are? i just looked at my notes and copied how it worked(i should take better notes).

    ok now that i filled in the triangle, now to sub it back for [tex]\theta[/tex]

    [tex] \frac{1}{16}(tan \theta) + \frac{1}{16}(sec \theta)^2 [/tex]
    will turn to...


    [tex] \frac{1}{16}(\frac{\sqrt{(4x^2-1)}}{x^3}) + \frac{1}{16}(\frac{x^3}{2})^2 [/tex]

    and that is my answer, is the correct?
     

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  9. Sep 11, 2004 #8
    The second integral should have solution tan³(x)/3, i think you made a mistake there.
     
  10. Sep 11, 2004 #9
    [tex] [\int sec^2\theta tan^2\theta d\theta] [/tex]

    is equal to tan³(theta)/3

    regards
    marlon
     
  11. Sep 11, 2004 #10
    i dont see how you got that, i did the problem agian and got the same answer
     
  12. Sep 11, 2004 #11

    Pyrrhus

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    Marlon is right, your first integral is correct but not your second.

    Let's work out the second integral with substitution..

    [tex] u=tan\theta [/tex]
    [tex] du=sec^2\theta d\theta [/tex]

    so we will have:
    [tex] \int u^2 du [/tex]

    Integrating:
    [tex] \frac{u^3}{3} + C [/tex]

    Substituting back:
    [tex] \frac{tan^3\theta}{3} + C [/tex]


    your result should be:

    [tex] \frac{tan\theta}{16} + \frac{tan^3\theta}{48} + C [/tex]

    Now to change it back:

    Look at what we had above:
    [tex] \sqrt{4x^2-1}= tan\theta [/tex]

    so
    [tex] \frac{\sqrt{4x^2-1}}{16} + \frac{(\sqrt{4x^2-1})^3}{48} + C [/tex]
     
    Last edited: Sep 11, 2004
  13. Sep 11, 2004 #12

    Pyrrhus

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    I just noticed this is a definite integral... oh well just apply the theorem :smile:
    and forget about the constant.
     
  14. Sep 11, 2004 #13
    sorry, you guys are right....


    well anyway, with the correct results...

    [tex] \frac{tan\theta}{16} + \frac{tan^3\theta}{48} + C [/tex]


    time to sub in theta...

    [tex] \frac{\frac{\sqrt{(4x^2-1)}}{x^3}}{16} + (\frac{\frac{\sqrt{(4x^2-1)}}{x^3})^3}{48} + C [/tex]

    is that right? i drew a triangle and everything in my eariler post


    Cyclovenom got a answer of [tex] \frac{\sqrt{4x^2-1}}{16} + \frac{(\sqrt{4x^2-1})^3}{48} + C [/tex]

    but i still have the x^3 in mines, what am i doing wrong? tan is o/a right? so i just subbed it in
     
    Last edited: Sep 11, 2004
  15. Sep 11, 2004 #14

    Pyrrhus

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    On your triangle you should have on:

    Hypotenuse: [tex] 2x [/tex]
    Opposite: [tex] \sqrt{4x^2-1} [/tex]
    Adjacent:[tex] 1 [/tex]
     
  16. Sep 11, 2004 #15

    sorry to keep on bothering you guys, but how did you get those values?
     
  17. Sep 11, 2004 #16

    Pyrrhus

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    Well, you started with [tex] 2x = sec\theta [/tex] so if you read above HallsofIvy post on how to get a value that will work for the [tex] \sqrt{4x^2-1} [/tex] which will be [tex] tan\theta [/tex]. You can build a Triangle with that info.

    You know [tex] sec\theta = \frac{Hypotenuse}{Adjacent}[/tex] and you got
    [tex] tan\theta = \frac{opposite}{adjacent} [/tex]

    so [tex] \frac{2x}{1} = \frac{Hypotenuse}{Adjacent}[/tex] and [tex] \frac{\sqrt{4x^2-1}}{1} = \frac{opposite}{adjacent} [/tex]

    Also:

    [tex] \frac{1}{cos\theta} = 2x[/tex]

    [tex] cos\theta = \frac{1}{2x}[/tex]

    and:

    [tex]\frac{\sqrt{4x^2-1}}{1} = \frac{sin\theta}{cos\theta}[/tex]

    [tex]\frac{\sqrt{4x^2-1}}{1}\frac{1}{2x} = \frac{sin\theta}{cos\theta}cos\theta[/tex]

    [tex]sin\theta = \frac{\sqrt{4x^2-1}}{2x} [/tex]

    [tex] \frac{Opposite}{Hypotenuse} = \frac{\sqrt{4x^2-1}}{2x} [/tex]

    Do you see it?
     
    Last edited: Sep 11, 2004
  18. Sep 11, 2004 #17
    yea!!! thank you, i get it.
     
  19. Sep 11, 2004 #18

    lol, i just figured out what you tried to tell me, im a bit slow. wow, that helped alot. my teacher goes really fast, and doesnt explain the little details and the book is just weird. i already solved 6 problems on my own(14 more to go) and got them all right. thank you agian, i love this forum
     
  20. Sep 11, 2004 #19

    Pyrrhus

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    Well, i'm glad i was of help :smile:
    Keep solving problems!! :cool:
    and feel free to ask for help anytime as long as you will learn from it.
     
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