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PLease help me in this integration problem

  1. May 31, 2005 #1
    PLease help me in solving the following integral!!


    please help me as soon as possible because i want to submit my assignment tomorrow
  2. jcsd
  3. May 31, 2005 #2
    [tex]\int{\sqrt{\tan x}dx}[/tex]

    I would imagine if you played around with it enough you could get it by parts. I cant see anything obvious though.
  4. May 31, 2005 #3
    \tan{x}=t \Rightarrow \frac{dx}{(\cos{x})^{2}}=dt[/tex], and remember [tex]\frac{1}{(\cos{x})^{2}}=1+(\tan{x})^{2}.[/tex]
  5. May 31, 2005 #4
    [tex] x = arctan(u) [/tex]

    [tex] dx = \frac{du}{1+u^2} [/tex]

    [tex] \int \sqrt{ \tan x} \ dx \Rightarrow \int \frac{\sqrt{ u }}{1+u^2} \ du [/tex]

    Can you explain the discrepancy? I cant.

    http://www.public.asu.edu/~hyousif/ma.JPG [Broken]
    Last edited by a moderator: May 2, 2017
  6. May 31, 2005 #5


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    There's no discrepancy whasoever.The 2 integrals are different.If you were to invert the sub.which you started with,the 2 results would coincide.

  7. May 31, 2005 #6
    Oh, ofcourse. How could I forget that.
  8. May 31, 2005 #7
    Since this is an indeterminate integral, you have to find the anti derivative of the function [tex] f(x) = \sqrt{\tanx} [/tex]

    I would do what whozum did, which I like to call "u-substitution"

    [tex] u = \tanx[/tex], so [tex]du = sec^2x dx[/tex] and [tex]dx = \frac{du}{sec^2x}[/tex]

    Is this right Whozum?

    Using what I got, [tex] dx = \cos^2x du[/tex]
    Can u work from here ?
    Last edited: May 31, 2005
  9. May 31, 2005 #8
    Yeah, you've derived the same result as I have, from there I think parts is the way to go.
  10. May 31, 2005 #9
    Please help me a little more

    Thanks a lot for ur precious help!

    friends! im still having a little problem in solving it. please can u people help me a little further to solve this integral.

    thanks in advance!
  11. May 31, 2005 #10
    I can't provide any real help to the answer of this integral, but I did do it online and I've attached the answer. It's long, but maybe it will provide some insight.


    Attached Files:

  12. May 31, 2005 #11
    Well it's not that difficult or ugly... After the substitution i pointed out it is simply:
    [tex]\int{\sqrt{t}(t^{2}+1)dt}=...=\frac{2}{7}(\tan{x})^{\frac{7}{2}}+\frac{2}{3}(\tan{x})^{\frac{3}{2}}+C [/tex]
  13. May 31, 2005 #12
    Oggy the final integral is

    [tex] \int \frac{\sqrt{ u }}{1+u^2} \ du [/tex]
  14. May 31, 2005 #13
    [tex] \int \frac{\sqrt{ u }}{1+u^2} \ du [/tex]

    [tex] g = \sqrt{u} \ \ \ h = arctan(u) [/tex]

    [tex] dg = \frac{1}{2\sqrt{u}}\ du \ \ \ dh = \frac{1}{1+u^2} du [/tex]

    [tex] \int g dh = gh - \int h dg [/tex]

    [tex] \int \frac{\sqrt u}{1+u^2} du = \sqrt{u}\ arctan u - \int \frac{arctan(u)}{2\sqrt{u}} du [/tex]
    Last edited: May 31, 2005
  15. Jun 1, 2005 #14
    Yep u^2+1 should be in the denominator. Sorry
  16. Jun 1, 2005 #15
    Thanks once again but what is its final answer

    Hi Guys

    thank you very much for ur tremendous help. You people are really very helpful.

    Now whozum!

    i have tried to solve this integral, from the point where u have left it. but im getting a zero answer.

    why is its so.

    should i try to solve it further from the point where u have left it or should i stop it there. if further solution is required then what will be the right answer so that i should get it.

    Please PLease help me! this question has now become a challenge for me.

    Thanks in advance!
  17. Jun 1, 2005 #16


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    Here's how it's done.

    [tex] I=\int\sqrt{\tan x} \ dx [/tex] (1)

    Assuming the tangent is defined on an interval on which is positive,so we don't run into complex valued functions,one makes fisrt sub

    [tex] \tan x=t^{2} [/tex] (2)

    [tex] dx=\frac{2t}{1+t^{4}} dt [/tex] (3)


    [tex] I=2\int \frac{t^{2}}{1+t^{4}} \ dt [/tex] (4)

    Use the trick

    [tex] 1+t^{4}=\left(1+t^{2}\right)^{2}-\left(\sqrt{2}t\right)^{2} [/tex] (5)

    ,so that (4) becomes

    [tex] I=2\int \frac{t^{2}}{\left(1+\sqrt{2}t+t^{2}\right)\left(1-\sqrt{2}t+t^{2}\right)} \ dx [/tex] (6)

    Use partial fractions and maybe partial integrations...

  18. Jun 1, 2005 #17


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    [tex] I=2\int \frac{t^{2}}{\left(1+\sqrt{2}t+t^{2}\right)\left(1 -\sqrt{2}t+t^{2}\right)} \ dx [/tex]

    Can someone check through Maple whether the answer to this is consistent with the original integral?
  19. Jun 1, 2005 #18


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    If you're doubting the calculations exposed above and,incidentally,not shortcut,why won't you put a pencil in your precious hand and grab a piece of paper and do it...?

    Sides,Maple is a known f***-up.:tongue2:

  20. Jun 1, 2005 #19
    I don't think that's right; I think it should be:
    [tex]\frac{-\sqrt{2}(\ln(\vert(\sqrt{2\tan (x)}+1)\cos (x) + \sin (x)\vert)-\ln(\vert(\sqrt{2\tan (x)}-1)\cos (x) - \sin(x)\vert)-2(\tan^{-1}(\sqrt{2\tan (x)}+1)+\tan^{-1}(\sqrt{2\tan(x)}-1)))}{4}[/tex]
  21. Jun 1, 2005 #20
    I don't think
    [tex] I=2\int \frac{t^{2}}{\left(1+\sqrt{2}t+t^{2}\right)\left(1 -\sqrt{2}t+t^{2}\right)} \ dx [/tex]
    is consistent with the original integral, because that would be [itex]-\cos^2 x[/itex]
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