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Integration Help

  1. Oct 4, 2013 #1
    1. The problem statement, all variables and given/known data

    ∫(9+16x2)3/2dx

    2. Relevant equations



    3. The attempt at a solution

    I have tried to solve this one in various ways. At first, I tried using trig substitutions and then integration by parts. I attached a pdf of what I tried to do. I'm unsure as to what I am doing wrong. Is there something I am missing here? I tried to do a u substitution with u=√9+16x2 but that seemed to not work either. Other ways just get me bogged down in a very long integration by parts thing. Is there any other way? Can anyone shed some light on what I can do to solve this? Thanks!
     

    Attached Files:

  2. jcsd
  3. Oct 4, 2013 #2

    jfizzix

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    It seems like trig substitutions are the way to go. I looked at it quickly, and if you can integrate [itex]sec^{5}(\theta)[/itex], you can solve the problem
     
  4. Oct 4, 2013 #3
    That's what I kept getting to. However, it seemed like it was going to take many integration by parts. I guess there is no other option. Any advice on how to integrate sec^5 (x). I saw somewhere else to write it as
    cos (x) / cos^6 (x)
     
  5. Oct 5, 2013 #4

    verty

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    You can do it by parts, it is ugly though.

    Let d(sec) = sec(x) tan(x) dx
    So int sec tan^2 dx = int tan d(sec)

    I'll use this to make the structure easier to see.

    sec^5 = sec (1 + tan^2)^2
    = sec + 2 sec tan^2 + sec tan^4

    int sec^5 dx = int sec dx + 2 int tan d(sec) + int tan^3 d(sec)


    Start with the most difficult term and work your way down.

    int tan^3 d(sec) = sec tan^3 - int sec 3 tan^2 sec^2 dx
    = sec tan^3 - 3 int sec^3 tan^2 dx
    = sec tan^3 - 3 int sec tan^2 (1 + tan^2) dx
    = sec tan^3 - 3 int tan d(sec) - 3 int tan^3 d(sec)

    This should give you enough confidence to continue in this direction and get the answer.
     
  6. Oct 5, 2013 #5

    haruspex

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    I think it's a little easier substituting sinh instead of tan. You get cosh4, which you can then handle by expanding as a sum of exponentials.
     
  7. Oct 5, 2013 #6
    Thing is, I have never learned about hyperbolic sines or cosines. In Calc 1 we were told to omit them, and haven't gone over them in Calc II. Although I hate doing this, I'm trying to do it by remembering the reduction formula for secant^m(x). Every time I do it by parts I seem to get lost in 2 pages of integrals.

    I think it's sec^m(x)=sinxsec^(m-1)/(m-1) + (m-2)/(m-1)int sec^(m-2) dx
     
    Last edited: Oct 6, 2013
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