Results 1 to 7 of 7

Thread: Integration

  1. #1
    Junior Member
    Join Date
    Aug 2007
    HSC
    2008
    Gender
    Male
    Posts
    194
    Rep Power
    6

    Integration

    Integration:

    I=integral:[lnsinx] boundaries a=pi/2 b=0

    nudge nudge, help?
    It takes courage to admit that you're wrong.

  2. #2
    Pimpcess.Snaz
    Guest

    Re: Integration

    This is a bitch.

    I tried to do it using the parts method where i made
    u = lnsinx
    du = cosx/sinx

    and dv = dx
    v = x

    Then i got I = xlnsinx - integral (xcosx/sinx)

    Then i made integral(xcosx/sinx) = I(2)

    u = x
    du = dx

    dv = sinx/cosx
    v=lnsinx

    I(2) = xlnxsinx - integral (lnsinx)
    I(2) = xlnsinx - I

    I = xlnsinx - (xlnsinx - I)
    I = xlnsinx - xlnsinx + I
    I = I


    lol i did that three times.. sorry i wasn't much help

  3. #3
    Yip
    Yip 当前离线
    Junior Member Yip's Avatar
    Join Date
    Sep 2005
    HSC
    2006
    Gender
    Male
    Posts
    140
    Rep Power
    7

    Re: Integration

    This integral is just an exercise in substitution lol
    I'll use the notation I[a,b]...dx to represent the definite integral from a to b wrtx.

    Let J=I[0,pi/2]log(sinx)dx
    Let x=pi/2-t in the integral
    dx=-dt
    I[0,pi/2]log(sinx)dx=I[pi/2,0]-log(cost)dt
    =I[0,pi/2]log(cost)dt=I[0,pi/2]log(cosx)dx
    (this can be seen from letting t=x. For the rest of this proof, I wont use the intermediate sub, but just write it out in terms of x, as the t sub is just a dummy variable)

    Since I[0,pi/2]log(sinx)dx=I[0,pi/2]log(cosx)dx,
    2J=I[0,pi/2]log(sinxcosx)dx
    =I[0,pi/2]log(0.5sin2x)dx
    =I[0,pi/2]log(0.5)dx+I[0,pi/2]log(sin2x)dx
    =(-pi/2)log2+I[0,pi/2]log(sin2x)dx
    =(-pi/2)log2+0.5I[0,pi]log(sinx)dx
    =(-pi/2)log2+0.5(I[0,pi/2]log(sinx)dx+I[pi/2,pi]log(sinx)dx)
    =(-pi/2)log2+0.5(J+I[0,pi/2]log(sinx)dx) (note:I[pi/2,pi]log(sinx)dx=I[0,pi/2]log(sinx)dx through the substitution x=pi-t)
    =(-pi/2)log2+J

    J=I[0,pi/2]log(sinx)dx=(-pi/2)log2
    Last edited by Yip; 14 Mar 2008 at 7:30 PM.

  4. #4
    boredofposting
    Join Date
    Jul 2006
    HSC
    2008
    Gender
    Male
    Location
    In your mum
    Posts
    105
    Rep Power
    6

    Re: Integration

    How did you do this:

    I[0,pi/2]log(0.5sin2x)dx=I[0,pi/2]log(0.5)dx+I[0,pi/2]sin2xdx

    shouldn't there be a log in front of sin 2x?


    Also, how did you do:

    (-pi/2)log2+I[0,pi/2]sin2xdx=(-pi/2)log2+0.5I[0,pi]sinxdx

    Probably some extremely clever trick that I don't know about


    I can see genius in that working out. Please explain
    I think the first example I posted is wrong though.

  5. #5
    Yip
    Yip 当前离线
    Junior Member Yip's Avatar
    Join Date
    Sep 2005
    HSC
    2006
    Gender
    Male
    Posts
    140
    Rep Power
    7

    Re: Integration

    Yeah, it was a typo on my part lol. I've edited the working accordingly.

    In I[0,pi/2]log(sin2x)dx, u let x=t/2
    dx=0.5dt
    I[0,pi/2]log(sin2x)dx=0.5I[0,pi]log(sint)dt=0.5I[0,pi]log(sinx)dx

  6. #6
    boredofposting
    Join Date
    Jul 2006
    HSC
    2008
    Gender
    Male
    Location
    In your mum
    Posts
    105
    Rep Power
    6

    Re: Integration

    Quote Originally Posted by 3unitz
    impossible to read...

    Where there's a will...

  7. #7
    boredofposting
    Join Date
    Jul 2006
    HSC
    2008
    Gender
    Male
    Location
    In your mum
    Posts
    105
    Rep Power
    6

    Re: Integration

    Quote Originally Posted by Yip
    This integral is just an exercise in substitution lol
    I'll use the notation I[a,b]...dx to represent the definite integral from a to b wrtx.

    Let J=I[0,pi/2]log(sinx)dx
    Let x=pi/2-t in the integral
    dx=-dt
    I[0,pi/2]log(sinx)dx=I[pi/2,0]-log(cost)dt
    =I[0,pi/2]log(cost)dt=I[0,pi/2]log(cosx)dx
    (this can be seen from letting t=x. For the rest of this proof, I wont use the intermediate sub, but just write it out in terms of x, as the t sub is just a dummy variable)

    Since I[0,pi/2]log(sinx)dx=I[0,pi/2]log(cosx)dx,
    2J=I[0,pi/2]log(sinxcosx)dx
    =I[0,pi/2]log(0.5sin2x)dx
    =I[0,pi/2]log(0.5)dx+I[0,pi/2]log(sin2x)dx
    =(-pi/2)log2+I[0,pi/2]log(sin2x)dx
    =(-pi/2)log2+0.5I[0,pi]log(sinx)dx
    =(-pi/2)log2+0.5(I[0,pi/2]log(sinx)dx+I[pi/2,pi]log(sinx)dx)
    =(-pi/2)log2+0.5(J+I[0,pi/2]log(sinx)dx) (note:I[pi/2,pi]log(sinx)dx=I[0,pi/2]log(sinx)dx through the substitution x=pi-t)
    =(-pi/2)log2+J

    J=I[0,pi/2]log(sinx)dx=(-pi/2)log2

    Fucking genius.


    Yip, how did you know to sub pi-t ? Did you just experiment, or is there a definitive method to it?


LinkBacks (?)

Thread Information

Users Browsing this Thread

There are currently 1 users browsing this thread. (0 members and 1 guests)

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •