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Integration (by parts)

  1. Oct 12, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\int_0^1 (6t^2 (1+9t^2)^{1/2} dt) [/tex]



    2. Relevant equations
    [tex]\int u dv = u v - \int v du [/tex]


    3. The attempt at a solution

    [tex]\int_0^1 (6t^2 (1+9t^2)^{1/2} dt) [/tex]
    [tex]=6 * \int (t^2 (1+9t^2)^{1/2} dt)[/tex]
    [tex]= 6 * \int (t * t (1+9t^2)^{1/2} dt)[/tex]

    Let [tex]u = t [/tex]; let [tex]dv = t (1+9t^2)^{1/2} dt [/tex];
    then [tex]du = dt[/tex]; and [tex]v = \int t (1+9t^2)^{1/2} dt[/tex]

    (using w-substitution:
    [tex] w = 1+9t^2[/tex],
    [tex] dw = 18t dt[/tex];
    [tex] dw/18=dt[/tex];
    [tex] \int t (1+9t^2)^{1/2} dt [/tex]
    =[tex] 1/18 \int w^{1/2} dw = 1/18 * 2/3 w^{3/2} = w^{3/2} / 27 = (1+9t^2)^{3/2}/27 [/tex]

    [tex] v = [(1+9t^2)^{3/2}/27 [/tex]

    [tex] \int(u dv) = u v - \int (v du) [/tex]
    [tex] = t * (1+9t^2)^{3/2}/27 - \int ((1+9t^2)^{3/2}/27 dt)[/tex]

    now i need help integrating [tex] (1+9t^2)^{3/2}/27 [/tex].
     
  2. jcsd
  3. Oct 12, 2008 #2

    tiny-tim

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    Hi Knissp! :smile:
    oooh, that's horrible! :cry:

    Hint: start again, and make the obvious substitution at the beginning. :wink:
     
  4. Oct 12, 2008 #3
    Can you use trig-sub or are you required to solve it by int. by parts?
     
  5. Oct 12, 2008 #4
    I can use any method, I just thought by parts would be easiest. What trig subst. would I use?

    and Tim: would the obvious substitution be u=t^2 and dv = sqrt(1+9t^2) dt? Then I have to integrate sqrt(1+9t^2) dt...
     
  6. Oct 12, 2008 #5
    [tex]t=3\tan\theta[/tex]
     
  7. Oct 12, 2008 #6
    [tex] t=3\tan\theta [/tex]
    [tex] dt = 3\sec^2\theta d\theta [/tex]

    [tex] \int (6t^2 (1+9t^2)^{1/2} dt) [/tex]
    = [tex] \int (6(3\tan\theta )^2 (1+9(3\tan\theta)^2)^{1/2} 3\sec^2\theta d\theta) [/tex]
    =[tex] \int (54 \tan^2\theta (1+81 \tan^2\theta)^{1/2} 3\sec^2\theta d\theta) [/tex]
    =[tex] 162 \int (\tan^2\theta (1+81 \tan^2\theta)^{1/2} \sec^2\theta d\theta) [/tex]

    Should I keep going from here?
     
  8. Oct 12, 2008 #7
    OMG!!! I'm so sorry ... lol

    First, you have to simplify it ... to find what your "a" should be.
     
  9. Oct 12, 2008 #8

    gabbagabbahey

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    hmm... I think rocomath meant to say [itex]3t=tan(\theta)[/itex] ;0)
     
  10. Oct 12, 2008 #9
    [tex]1+9t^2[/tex]

    Dividing by 9, [tex]\sqrt{9\left(\frac 1 9+t^2\right)}=3\sqrt{\frac 1 9+t^2}[/tex]

    [tex]t=\frac 1 3\tan\theta[/tex]
     
  11. Oct 12, 2008 #10
    Now I have
    [tex] \int 2/9 \tan^2\theta \sec^3\theta d\theta [/tex]

    which is

    [tex] \int 2/9 \frac{\sin^2\theta}{\cos^5\theta} d\theta [/tex]
     
  12. Oct 12, 2008 #11

    gabbagabbahey

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    Try Integration by parts now (don't forget that your limits of integration are [itex]tan^{-1}(0) \rightarrow tan^{-1}(3)[/itex] now)
     
  13. Oct 12, 2008 #12

    tiny-tim

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    Hi Knissp!

    (have a theta: θ and a squared: ² and a cubed: ³ and an integral: ∫ :smile:)

    Isn't it tan²θsec³θ?
     
  14. Oct 12, 2008 #13
    I have tried several ways none of which worked, too much to post it all. Just a little more help needed: would it be best to use the identity 1+tan²θ = sec²θ before integrating by parts?


    edit: just to add, in case anyone wanted to know where this problem is from,

    Let C be the curve represented by the equations
    x=2t, y=3t² (0≤t≤1) dx/dt=2, dy/dt=6t
    Evaluate the line integral along C: ∫(x-y)ds
    ds = √((dx/dt)^2+(dy/dt)^2) dt = √(4+36t^2) = 2√(1+9t^2) dt
    ∫(x-y)ds = ∫((2t-3t^2)*2√(1+9t^2) dt = ∫((4t√(1+9t^2) dt - ∫((6t^2)√(1+9t^2) dt
    I was able to solve the ∫((4t√(1+9t^2) dt part using u-substitution, but I got stuck on ∫((6t^2)√(1+9t^2) dt which is why I originally asked for help. Question: Is my way of attempting this problem the most efficient, or is there a better way so I can avoid this integral which involves trig-substitution entirely?
     
    Last edited: Oct 12, 2008
  15. Oct 12, 2008 #14

    gabbagabbahey

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    Use the Identity [itex]sin^2(\theta)=1-cos^2(\theta)[/itex] to break the integral into two parts (sec^3 and sec^5). Then use integration by parts with [itex]u=sec(\theta)[/itex] and [itex]dv=sec^2(\theta)d\theta[/itex]....you will have to use by parts twice for the sec^5 term....you'll also need to look up the integral of [itex]sec(\theta)d\theta[/itex]
     
  16. Oct 12, 2008 #15
    tan²θsec³θ = (sec²θ-1)sec³θ = [tex]\sec^5\theta - \sec^3\theta[/tex]
    Looking at the sec^3 part, let u = secθ, dv=sec²θ dθ; du=secθtanθ, v=tanθ.
    uv-∫vdu = secθtanθ - ∫tan²θsecθ dθ
    Is that right so far? because then I have to make another u-sub...
     
  17. Oct 12, 2008 #16

    gabbagabbahey

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    looks fine so far; now there is a trick you need to use...
    -∫tan²θsecθ dθ =∫(1-sec²θ)secθ dθ =∫secθ dθ-∫secθ^3 dθ
    => 2 ∫secθ^3 dθ=secθtanθ+∫secθ dθ
     
  18. Oct 12, 2008 #17
    Aaahhh I get it now. :bugeye: Sorry that so long. Thank you all for your help!
     
  19. Oct 13, 2008 #18

    tiny-tim

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    hyperbolic trig substitution

    hmm … when you have a square, there's always the choice of a trig substitution, or a hyperbolic trig substitution …

    in this case, either t = (1/3)tanθ, or t = (1/3)sinhu …

    using the tan worked, but it was rather complicated.

    Try it again, using the sinh instead! :smile:
     
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