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Homework Help: Help with integration

  1. Feb 5, 2005 #1
    A have to integrate x^4/Sqrt((1-x^2)^3)

    I tried x = Sint, but it didn't help me.
    What kind of substitution I should use?
     
  2. jcsd
  3. Feb 5, 2005 #2

    dextercioby

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    Yes,it can help u.4 times partial integration + watch oiut woth the signs and u'll get the result.

    Daniel.
     
  4. Feb 5, 2005 #3
    Hm.
    I don't see even how to integrate dx/Sqrt((1-x^2)^3). (If i correctly understood Your idea)
    Tehere should be any substitution
     
  5. Feb 5, 2005 #4

    dextercioby

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    Yes,the "sine" substitution will get u
    [tex] \int \frac{\sin^{4}t}{\cos^{5}t} dt [/tex]

    ,which can be solved via partial integration & simple fractions.

    Daniel.

    EDIT:DISREGARD THIS MESSAGE,PLEASE!!!!!!! :redface:
     
    Last edited: Feb 5, 2005
  6. Feb 5, 2005 #5
    No. "Sine" substitution gave me (Sint)^4/(Cost)^2
    x=Sint
    dx=Cost dt
     
    Last edited: Feb 5, 2005
  7. Feb 5, 2005 #6

    dextercioby

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    What??
    [tex] (1-x^{2})^{3}\rightarrow \cos^{6}t [/tex] agree??

    [tex] dx\rightarrow \cos t \ dt[/tex]

    [tex] x^{4}\rightarrow \sin^{4}t [/tex]

    Make the proper ratio & u'll see that i was right.

    Daniel.
     
    Last edited: Feb 5, 2005
  8. Feb 5, 2005 #7

    arildno

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    [tex]x=sin(t)[/tex]
    [tex]\frac{x^{4}}{\sqrt{(1-x^{2})^{3}}}=\frac{\sin^{4}{t}}{\sqrt{\cos^{6}t}}=\frac{\sin^{4}t}{|\cos^{3}t|}[/tex]
     
  9. Feb 5, 2005 #8

    dextercioby

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    UUUUUUUUUPS,i didn't see the sqrt... :blushing:

    Daniel;
     
  10. Feb 5, 2005 #9
    Arildno, i think you have forgotten dx= Cost dt
     
  11. Feb 5, 2005 #10

    dextercioby

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    He hasn't he didn't deal with the "dx",just with the ratio...

    Daniel.
     
  12. Feb 5, 2005 #11

    arildno

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    The one thing missing in your account, is the absolute value sign.
    You should end up with:
    [tex]\int{sign}(\cos{t})\frac{\sin^{4}t}{\cos^{2}t}dt[/tex]
     
  13. Feb 5, 2005 #12

    dextercioby

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    Anyway,the advice with partial integration is still valid.U'll need the double angle formula,though.

    Daniel.
     
  14. Feb 5, 2005 #13
    I think I do not need partial integration. I can (Cost)^2=1-(Sint)^2 and then divide.
    There are rather simple integrals.

    But I realy don't understand why I must take into account sign(cost).
    I have never seen anything similar
     
  15. Feb 5, 2005 #14

    arildno

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    Alternatively, you may rewrite this expression in terms of the tangent function.
     
  16. Feb 5, 2005 #15

    arildno

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    Well, you'll get wrong results if you don't.

    It doesn't provide you with anymore trouble, though:
    In regions when cos(t)>0, sign(cos(t))=1, whereas when cos(t)<0, sign(cos(t))=-1
     
  17. Feb 5, 2005 #16
    I finally got the result. And it is correct! Thank You all.
     
  18. Feb 5, 2005 #17

    dextercioby

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    Yes,Arildo,things are much more simple that they were.U've forgotten where he started from
    [tex] \sqrt{(1-x^{2})^{3}} [/tex]

    into reals has sense only for [itex] -1<x<+1 [/tex]

    Making the substitution
    [tex] x\rightarrow \sin t [/tex]

    "t" must be in this interval and other (dictated by the periodicity of "sine")
    [tex] -\frac{\pi}{2}<t<+\frac{\pi}{2} [/itex]

    where the cosine is strictly POSITIVE.No need for modulus... :wink: It can be proven that in all other real intervals where "sine t" obeys the inequality
    [tex] -1<\sin t<+1 [/tex]

    the cosine is positive strictly.

    Daniel.
     
  19. Feb 5, 2005 #18

    arildno

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    Oh, dear..the shame of it..:redface: (goes back to my bucket)
     
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