1. Limited time only! Sign up for a free 30min personal 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!

Integration help.

  1. May 21, 2009 #1

    trv

    User Avatar

    1. The problem statement, all variables and given/known data

    Need to integrate

    dt=da (8*pi*rho0/3a-1)^-1/2

    to get

    t=(8*pi*rho0/3)*arctan(1/sqrt(8*pi*rho0/3a-1))-sqrt(a)*sqrt(8*pi*rho0/3-a)


    2. Relevant equations

    3. The attempt at a solution
    I'm totally lost. I'm guessing there's substitution involved, but unsure on how to get the arctan in. Any thoughts. If someone can tell me how to get the arctan in I think I'll be able to give a better go at it.

    Thanks in advance.
     
    Last edited: May 21, 2009
  2. jcsd
  3. May 21, 2009 #2

    Cyosis

    User Avatar
    Homework Helper

    Could you list the entire problem, because this doesn't make sense. You say you need to integrate [itex]4\pi r^2[/itex]. The only variable in there is r, does r depend on t, what are you integrating over, what does it mean? The answer has a [itex]\rho_0[/itex] and an a, yet the original function does not. There is a template for a reason. List the full problem, list all variables including their meaning.
     
  4. May 21, 2009 #3

    trv

    User Avatar

    hey sorry, i cant seem to get the latex right. Let me try again

    need to integrate

    dt=da (8*pi*rho0/3a-1)^-1/2

    Its a separable equation in a and t. the rhs being integrated wrt a

    the solution should be
    t=(8*pi*rho0/3)*arctan(1/sqrt(8*pi*rho0/3a-1))-sqrt(a)*sqrt(8*pi*rho0/3-a)
     
    Last edited: May 21, 2009
  5. May 21, 2009 #4

    Cyosis

    User Avatar
    Homework Helper

    This will not yield an arctan. I still don't really know what you're trying to calculate.

    Also what is [itex]A_n[/itex]?
     
  6. May 21, 2009 #5

    trv

    User Avatar

    :(.

    I'm working through my cosmology notes. It's an integration that comes up while solving the Friedmann equations for the matter dominated cosmos with spatial curvature, k=+1.

    So it definitely can't be integrated to get arctan?

    Edit:Oh sorry i make a mistake. Please check it again. It should be -1/2 instead.
     
    Last edited: May 21, 2009
  7. May 21, 2009 #6

    Cyosis

    User Avatar
    Homework Helper

    This is the equation you want to integrate as I interpret it.

    [tex]dt=da \frac{1}{\frac{8 \pi \rho_0}{3a}-1}[/tex]. Is this what you had in mind?

    The Friedman equation for a matter dominated universe with k=+1 is given by:

    [tex]
    \left(\frac{\dot{a}}{a}\right)^2+\frac{c^2}{a^2}=\frac{\Omega_m}{a^3}[/tex]

    Is this what you're trying to solve?
     
    Last edited: May 21, 2009
  8. May 21, 2009 #7

    trv

    User Avatar

    Close but with the square-root sign around, (8*pi*rho0/3a-1)

    i.e. ^-1/2 rather than ^-1

    and, yes, u've got the right equation, with omega_m=8*pi*rho_0/3, and c^2=1.
     
  9. May 21, 2009 #8

    Cyosis

    User Avatar
    Homework Helper

    Ah the square root, now it finally makes sense yes you will get an arctan in it now. I will have a look at it.
     
  10. May 21, 2009 #9

    trv

    User Avatar

    Thanks a lot for ur patience.
     
  11. May 21, 2009 #10

    Cyosis

    User Avatar
    Homework Helper

    To reduce the amount of constants in our integral we're going to integrate:

    [tex]\int \frac{da}{\sqrt{\frac{\Omega}{a}-1}}[/tex]

    You can substitute the values for [itex]\Omega[/itex] back into the equation later.

    Start with substituting [itex]u=\frac{\Omega}{a}-1[/itex]. Note there will be multiple substitutions involved, but lets start here.
     
  12. May 21, 2009 #11

    trv

    User Avatar

    This should go to

    [itex]
    -\int \frac{\Omega du}{(u+1)^2\sqrt{u}}
    [/itex]

    if I haven't made a mistake. Then I guess integration by parts. I'll give that a go now.
     
  13. May 21, 2009 #12

    Cyosis

    User Avatar
    Homework Helper

    Correct, now make a substitution [itex]x=\sqrt{u}[/itex]. The next step is a trigonometric substitution. Remember [itex]\tan^2 x+1=\sec^2 x[/itex].
     
  14. May 21, 2009 #13

    trv

    User Avatar

    Oh ok.
     
  15. May 21, 2009 #14

    trv

    User Avatar

    [itex]
    -\int \frac{2\Omega dx}{(x^2+1)^2}
    [/itex]

    Is that right?
     
  16. May 21, 2009 #15

    Cyosis

    User Avatar
    Homework Helper

    Still correct. Can you see what kind of trig substitution to use now?
     
  17. May 21, 2009 #16

    trv

    User Avatar

    Does it have to be a trig. substitution? I dont know a trig. function that integrates to arctan x :(. I do know 1/(1+x^2) integrates to arctan x, so I can see we are quite close however.
     
  18. May 21, 2009 #17

    Cyosis

    User Avatar
    Homework Helper

    Well at first sight you may think you're close, because it looks so much like the derivative on of an arctan, but the square makes quite a difference. Substitute [itex]x=\tan s[/itex] and use the trigonometric identity I listed in one of my earlier posts.
     
  19. May 21, 2009 #18

    trv

    User Avatar

    [itex]
    -\int \frac{2\Omega ds}{sec^3s}
    [/itex]

    I think.


    Edit: dx was a typo. Changed it to ds.
     
    Last edited: May 21, 2009
  20. May 21, 2009 #19

    Cyosis

    User Avatar
    Homework Helper

    You have made an error somewhere, although you're close. Show me what you've done. Also why is there an dx in there and not a ds?
     
  21. May 21, 2009 #20

    trv

    User Avatar

    Think I found the mistake. Is it (sec s)^2 rather than (sec s)^3 in the equation above?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integration help.
  1. Integral help (Replies: 4)

  2. Integration Help (Replies: 5)

  3. Integration help! (Replies: 3)

  4. Help with an integral (Replies: 1)

  5. Integrals help (Replies: 7)

Loading...