1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    [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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

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

    Daniel.
     
  12. Feb 5, 2005 #11

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

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

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Oh, dear..the shame of it..:redface: (goes back to my bucket)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Help with integration
  1. Integral Help (Replies: 3)

  2. Help with Integral (Replies: 3)

  3. Integration Help (Replies: 2)

  4. Help with Integration (Replies: 1)

  5. Help integrating (Replies: 4)

Loading...