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Integragtion Problem with Substitution

  1. Feb 12, 2004 #1
    At the moment I'm trying to solve this integration problem and it's not working out asd neatly as any other substituion problem I've tried before.

    We have to integrate sqrtx divided by (1 + x^2) using the substitution x= tan^2t (as in tan squared t). So when we have dx/dt so you get 2tantsec^2t? I've tried using that but it all gets really muddled and complex. At the moment I'm down to finding the integral of 8 divided by -sin^2t but I'm not sue where to go from there. I've tried using substituions from this point on using the dpuble angle formulae but it's all going a bit pear-shaped. Any help would be much appreciated- thanks in advance!
  2. jcsd
  3. Feb 12, 2004 #2
    The integral of 1/sin^2(x) is -cot(x). Seems like the answer would be more complex...
    Last edited: Feb 12, 2004
  4. Feb 12, 2004 #3
    The answer we're supposed to get to is 2(sqrtx) - 2tan^-1(sqrtx)
  5. Feb 12, 2004 #4
    That's the integral of sqrt(x) / (x + 1) (notice the lack of a squared term in the denominator), which is "somewhat" easier than sqrt(x) / (x^2 + 1) ;)

    We wish to integrate sqrt(x) / (1 + x) dx.

    Let x = tan^2(t), which gives dx = 2tan(t)sec^2(t) dt.

    The integral turns into sqrt(tan^2(t)) / (1 + tan^2(t)) * 2tan(t)sec^2(t) dt. The numerator is equal to sec^2(t) (standard identity). We can simplify it to:

    tan(t)/sec^2(t) * 2tan(t)sec^2(t) dt =
    tan(t)2tan(t) dt =
    2tan^2(t) dt

    Hopefully you can handle the rest.
    Last edited: Feb 12, 2004
  6. Feb 12, 2004 #5
    Really? Eeek what's wrong with this sheet then? It tells us to use the substitution x=tan^2(x) so would we not then need this for the one you've suggested? V.confusing.
  7. Feb 12, 2004 #6
    Added more info to the last post :wink:

    Btw, try inputting Sqrt[x] / (x^2 + 1) at http://integrals.wolfram.com/ , and see what you get... Not the kind of thing you'd want work out for yourself on paper
    Last edited: Feb 12, 2004
  8. Feb 12, 2004 #7
    LOL, while you were typing that I was off tryingit myself and I've got down to the 2tan^2(t) bit! After that is there an integral you can use or do you have to change it to sin^2(x) over cos^2(x) and then use the double angle formulae? It just looks messy everytime I work with it? God, I should really remember to bring my brain home with me from uni!
  9. Feb 12, 2004 #8
    Personally, I'd rewrite it as tan^2(x) = (1 - cos^2(x)) / cos^2(x) = 1/cos^2(x) - 1 and work from there.
  10. Feb 12, 2004 #9

    matt grime

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    looks like a simple misprint, just use x=tan(t)

    for questions like this, try and fiddle cos^2 + sin^2 =1 so you get something that looks like what you want
  11. Feb 12, 2004 #10
    Hm okay, so if we had 1 over cos^2x - 1 would you then have to change that to -sin^x? You gave the integral of this earlier, well the integral of 1 over sin^2x which was -cotx, but how did you work that out? I'm generally fine with differentiating but integration is really confusing me, esp. when you have 1 over a trigonometric function.
  12. Feb 12, 2004 #11
    Looked it up :wink: I'm not aware of any way to "construct" such trigonometric integrals, other than simply checking the derivatives of the standard trig. functions (i.e sec(x), sec^2(x), csc(x), etc) and seeing if something fits.
  13. Feb 12, 2004 #12
    Hm but how do we get the correct answer then if we have the integral of 1 over -sin^2x?
  14. Feb 12, 2004 #13
    Okay if we end up with the integral being -cot(t) then what do we do? How do we get x back into it from what we made x equal to?
  15. Feb 12, 2004 #14
    I end up with 1/x. Odd.
  16. Feb 12, 2004 #15

    matt grime

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    Re: Re: Integragtion Problem with Substitution

    ignore me, didn't see the sqrt x in there

    int 1/sin^2(t) is cot(t) give or take a minus sign.


    so tan(t)=sqrt(x)

    and tan is 1 over cot

    so cot (t) is 1/sqrt(x)
  17. Feb 12, 2004 #16
    Okay, I just need to know how you get from the integral of 1 over -sin^t to 2sqrtx - 2tan^-1(sqrtx). It just doesn't seem to work.
  18. Feb 12, 2004 #17
    Looks like there's some confusion here. I said that tan^2(x) = 1/cos^2(x) - 1. The -1 is /not/ in the nominator. I meant (1/cos^2(x)) - 1, or sec^2(x) - 1. The integral of this is tan(x) - x (found a table of integrals here: http://www.math2.org/math/integrals/tableof.htm), so the integral of 2tan^2(t) dt (please excuse the sudden change of variables) is 2tan(t) - 2t.

    Since we let x = tan^2(t), get that that sqrt(x) = tan(t), or t = atan(sqrt(x)).

    Which gives 2tan(t) - 2t = 2tan(atan(sqrt(x))) - 2atan(sqrt(x)) = 2sqrt(x) - 2atan(sqrt(x)).

    Plus a constant. ;)
    Last edited: Feb 12, 2004
  19. Feb 12, 2004 #18

    matt grime

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    dunno if i'm helping or not, but, muzza, you've inconsistently muddled up t's and x's throughout these posts. perhaps claire starting again from the beginning would help along with her posting all her working out up to the point she gets stuck.
  20. Feb 12, 2004 #19
  21. Feb 12, 2004 #20
    Okay, I'll go scan in my working out because it's easier to see on paper!
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