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Homework Help: Integrating substitution problem?

  1. Dec 7, 2007 #1


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    [SOLVED] Integrating substitution problem?

    1. The problem statement, all variables and given/known data
    Sorry to hijack this thread sorta (as a similar named one already exists), but the title is aptly suited to my question.

    I have integral to integrete and I don't really know how to do it tbh. . .

    [tex] s=\int{\sqrt{2+(3t)^2}dt [/tex] Limits are from t=0 to t=5 (how do I show this on latek?)

    If it was 2-3t then I'd simply substitute the t for sin u, but what do I do when it positive?
    Last edited: Dec 7, 2007
  2. jcsd
  3. Dec 7, 2007 #2
    you want to get the radican in the form of a+bt^2, where b=1


    to show your limits on your integrand ... \int_{0}^{5}



    since your last problem was negative, you were able to use sine. this is negative, so what trigonometric identities do you know? options are secant and tangent.
    Last edited: Dec 7, 2007
  4. Dec 7, 2007 #3


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    Ahh nice... (still stuck) thanks a million and for the latek help too:D

    Looking at the trigfunctions .. .1 min . .
    Last edited: Dec 7, 2007
  5. Dec 7, 2007 #4
    you will need to do a little more manipulation. anytime!
  6. Dec 7, 2007 #5


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    Okkk I'm still stuck here..., if i take either sec or tan i'll get positives and I think i want to change it to a minus to the form 1-cos^2 (t) or sin the form. Or do I need to get this ... (you this is what happens when you develop a reliance upon the same thing working each time :redface:)
  7. Dec 7, 2007 #6
    well between secant or tangent, you want tangent b/c it gives you a positive in the radican


    the only thing you have to do is get it in the form of 1+tan^2(x)

    a little algebra


    still the same, so next let ...


    now take the derivative and substitute where it is appropriate.
    Last edited: Dec 7, 2007
  8. Dec 7, 2007 #7


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    Thanks very much for sticking with me on this :)

    i've got it down to here [tex] \sqrt{2}\int_{t=0}^{t=5}\sqrt{\frac{2}{9}(1+\tan^2 \theta)}\sec^2\theta \hspace{4} d\theta [/tex] ...
    hope that is right...
    ok so the sec will cancel giving sec^3 I'm assuming i just let [tex] u=\sec {\theta} [/tex]

    .. (still on it) ..
  9. Dec 7, 2007 #8
    you have everything correct so far except the constant inside the radical, it should be

    [tex]s=\sqrt{2}\int_{0}^{5}\sqrt{(\frac{\sqrt{2}}{3})^{2}[1+\tan^{2}\theta]}\sec^{2}\theta d\theta[/tex]

    which becomes

    [tex]s=\frac{2}{3}\int_{0}^{5}\sqrt{1+\tan^{2}\theta}\sec^{2}\theta d\theta[/tex]

    from there, you cannot just use a simple u=substitution, you will need to use Integration by Parts
  10. Dec 7, 2007 #9


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    I was wondering what on earth I should do to get the sec and tan out of there!!
    Ahh ok... woah this is cool integral.. *thanks the heavens this isn't his exam*
  11. Dec 7, 2007 #10
    lol, it shocked me too! i was like no way, wtf? haha, i really hate integrating secants, such a pain, but i'm with you to the end!
  12. Dec 7, 2007 #11


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    Ok, just before I start can I just confirm this is the sensible thing to do:

    I'm gonna let u=sec (theta) and dv = sec^2 (theta) d(theta)

    not sure if this is right the order, but my past experiences with these buggers is that the exponents are usually rarely let equal u if there is an alegrbaic or trig function present .. am I right?
  13. Dec 7, 2007 #12
    correct! make sure you keep writing your integrand on the right side = to your work


    [tex]\frac{2}{3}\int_{0}^{5}\sec^{3}\theta d\theta = \mbox{work done}[/tex]
  14. Dec 7, 2007 #13


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    THis is gonna be a long typing . ..

    Um, I'm getting an integral that seems to be going on to infinity, have done 3 int by parts and i'm stuck now with the integral of ln sec.sec^2 d(theta) .. . .wait!... nah still stuck

    [tex] let \hspace {4} u=\sec {\theta} [/tex] and [tex]dv=\sec^2 {\theta} d\theta}[/tex]

    This gives

    [tex]\frac{2}{3}\int_{0}^{5}\sec^{3}\theta d\theta =(\sec {\theta})*(\int{\sec^2 {\theta} d\theta}) - . .. . . . [/tex]

    Need to work out integral of \sec^2 {\theta} d\theta.

    Let [tex] k=\sec {\theta} [/tex] ... .wait should i let sec^2 theta equal 1/cos^2 (theta)??
    Last edited: Dec 7, 2007
  15. Dec 7, 2007 #14
    you only had to do parts once. after the first IbP, you use a trig substitution which gives you your original integrand and from there you just take the integral of secant.
  16. Dec 7, 2007 #15


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    Urgh ...well initially I did four of them .. .:redface:
  17. Dec 7, 2007 #16
    is all good :-]

    i was once given secant to the 5th to integrate and it took me a month. little mistakes and blind vision, i feel your pain.
  18. Dec 7, 2007 #17


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    Gahhh... this thing still ain't working out.. I cannot see any resemblance from the left to the right side, i keep getting the integral in terms of ln |sec| .

    [tex]\int_{t=0}^{t=5}\sec^3 {\theta} d\theta= \sec{\theta}\int{\sec^2\theta}d{\theta} .. . [/tex]

    Done this ,v part, integral numerous ways now,
    let sec^2 =1 +tan 2

    didn't work

    tried doing it by parts again and again
    latest effort

    let sec^2 = 1/cos^2
    and when i integrated that i got ln|sec (theta):yuck:
    Last edited: Dec 7, 2007
  19. Dec 7, 2007 #18
    [tex]\frac{2}{3}\int_{0}^{5}\sec^{3}\theta d\theta=\sec\theta\tan\theta|_{0}^{5}-\int_{0}^{5}\sec\theta\tan^{2}\theta d\theta[/tex]

    now use a trig identity to change tan^2(x) into secant and you will get your original integrand.

    now at this step, be careful!!! leave out your fraction of 2/3 for the next few steps. this is something i learned from secant to the 5th and kicked my ass all over the place.

    [tex]\frac{2}{3}[\int_{0}^{5}\sec^{3}\theta d\theta=\sec\theta\tan\theta|_{0}^{5}-\int_{0}^{5}\sec^{3}\theta+\int_{0}^{5}\sec\theta][/tex]

    now add secant cubed to the left side and you will get

    [tex]\frac{2}{3}[2\int_{0}^{5}\sec^{3}\theta d\theta=\sec\theta\tan\theta|_{0}^{5}-\int_{0}^{5}\sec\theta d\theta][/tex]
    Last edited: Dec 7, 2007
  20. Dec 7, 2007 #19


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    em how did you get v = tan (theta)

    When I did it I had u=sec (theta)

    meaning [tex]v= \int{\sec^2{\theta} d\{theta}}[/tex] This kept giving me the ln|sec| or and integral that went on forever, .. 1 sec.. .
    would you mind check to see where i'm going wrong
  21. Dec 7, 2007 #20
    your substitutions were

    [tex]du=\sec\theta\tan\theta d\theta[/tex]

    i think this is where you messed up

    [tex]dV=\sec^{2}\theta d\theta[/tex]

    dV also means

    [tex]dV=\int\sec^{2}\theta d\theta[/tex]


    [tex]I=uV-\int Vdu[/tex]
    Last edited: Dec 7, 2007
  22. Dec 7, 2007 #21


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    Oh right, this may sound trivial but I could never get sec^2 integrated = tan x .. i just kept getting loops :cry:
  23. Dec 7, 2007 #22


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    You know the derivative of tan is sec2. So it's just the fundamental theorem of calculus to go the other way
  24. Dec 7, 2007 #23


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    That was just .... grr.. .. i kept seeing the integral thing on the math tables (hence the ln's)... never looked at the differentiation things on the tables ......grrr... that's just the icing on the cake ..:mad:

    So simple!
    So bloody simple gah .. .how come I couldn't see that ... (lol i even googled abstract trigonometric identies and integrals) .. .

    Ok .. .give me a few more mins at this.. *must calm down first*
  25. Dec 7, 2007 #24
    so how's the integration going? lol
  26. Dec 7, 2007 #25


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    .. . ..

    [tex] I=\frac{2}{3} \int_{t=0}^{t5}sec^3(\theta) d\theta[/tex]


    [tex] \frac {5}{3}I=\sec\theta\tan\theta|_{t=0}^{t=5}+\ln{\sec(\theta)+\tan(\theta)}|_{t=0}^{t=5} [/tex] .. . .

    . . .
    [tex] t= \frac {\sqrt{2}}{3}\tan {\theta} [/tex]

    ==> [tex]\theta= tan^-1{\frac{3t}{\sqrt{2}} [/tex]

    @t=0 tan0 =0

    @t=5 tan^-1 (15/sqrt2) =84ยบ


    evaluated the new limits, and got the total arc length to be 52.84, and I think I'm gonna accept it.. now for the fun part as I've no answer to compare to I can't tell if it right but anyways, with the exception of evaluating the limits is the rest ok . .

    Thanks a million for the help Roco I really really appreciate it:D
    Last edited: Dec 7, 2007
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