[HotSnoJ]

http://www.clickherefree.com/cgi-bin/webdata_FreeWebHosting.pl?cgifunction=form&fid=1048029535


ROTFL

[Ashrak]

oooh

[Ulundel]

Wowsers...

[Tiara]

Think of the possibilities you will have with1 MB!:blah:

[Fetty]

Forced Ads: Banner/Top  

[Stealth]

i guess if you want a 2 page HTML website that's ok

[Carl]

though if you want to make an art gallery, this may be the place. you could fit a dozen jpgs or so in there.

[Gortef]

w00t

[Stealth]

Originally posted by Carl
though if you want to make an art gallery, this may be the place. you could fit a dozen jpgs or so in there.

yeah, a dozen 40x40 pixel JPEGs

[Petrarch of the VBB]

What. Is. The. Point.

[Tiara]

Code: [Select]

dy/dx = ay - by^3
    dy/dx - ay = -by^3    Divide through by y^3
  (1/y^3)dy/dx) - a/y^2 = -b

Make the substitution u = 1/y^2  So du/dx = (-2/y^3)(dy/dx)
                                  (-1/2)du/dx = (1/y^3)dy/dx)

Substitute for y and dy/dx

     (-1/2)du/dx) - au = -b
           du/dx + 2au = 2b

Multiply by the integrating factor e^(INT(2a.dx))
                                = e^(2ax)

 e^(2ax).du/dx + 2au.e^(2ax) = 2b.e^(2ax)

    d/dx{u.e^(2ax)} = 2b.e^(2ax)

Integrate   u.e^(2ax) = 2b.INT{e^(2ax).dx}
            u.e^(2ax) = 2b(1/(2a)).e^(2ax) + const.

      (1/y^2).e^(2ax) = (b/a).e^(2ax) + const


Thats the point... Get it now? :wink:

[Martinus]

Originally posted by Tiara

Code: [Select]

dy/dx = ay - by^3
    dy/dx - ay = -by^3    Divide through by y^3
  (1/y^3)dy/dx) - a/y^2 = -b

Make the substitution u = 1/y^2  So du/dx = (-2/y^3)(dy/dx)
                                  (-1/2)du/dx = (1/y^3)dy/dx)

Substitute for y and dy/dx

     (-1/2)du/dx) - au = -b
           du/dx + 2au = 2b

Multiply by the integrating factor e^(INT(2a.dx))
                                = e^(2ax)

 e^(2ax).du/dx + 2au.e^(2ax) = 2b.e^(2ax)

    d/dx{u.e^(2ax)} = 2b.e^(2ax)

Integrate   u.e^(2ax) = 2b.INT{e^(2ax).dx}
            u.e^(2ax) = 2b(1/(2a)).e^(2ax) + const.

      (1/y^2).e^(2ax) = (b/a).e^(2ax) + const


Thats the point... Get it now? :wink: [/B]

[color=66ff00]Ok CP! What did you do with Tiara's body?
[/color]

[an0n]

Methinks you phrased that wrong.

[Martinus]

Originally posted by an0n
Methinks you phrased that wrong.

[color=66ff00]Methinks you have a twisted imagination.

Considering CP's stance on girls how can you think of anything in that ballpark?
[/color]

[WMCoolmon]

This is getting out of hand...now there are two of them!

[DragonClaw]

Looks just like what I was looking for....  

[Carl]

Originally posted by Stealth


yeah, a dozen 40x40 pixel JPEGs

you can get a 800x600 to 80kb on good compression.

[Tiara]

...

Why is it that you guys think nonly CP knows his math?

z = tan(x/2).

Then

     sin(x) = 2*z/(1+z^2)
     cos(x) = (1-z^2)/(1+z^2)
         dx = 2*dz/(1+z^2)

Substitute this in, and you'll end up needing to integrate a rational function of z:

       INTEGRAL dx/(3+4*sin

• )^2

     = INTEGRAL 1/(3+4*2*z/[1+z^2])^2 * 2*dz/(1+z^2)
     = INTEGRAL (1+z^2)^2/(3*z^2+8*z+3)^2 * 2*dz/(1+z^2)
     = INTEGRAL 2*(1+z^2)/(3*z^2+8*z+3)^2 * dz

Followed by;

     3*z^2 + 8*z + 3 = 3*(z + [4+sqrt(7)]/3)*(z + [4-sqrt(7)]/3)

and partial fractions to break it into parts which you can easily
integrate.

[CP5670]

what was the original problem in that last quote?

try this one:

¥
ò ( e^(1-x)t - 1 ) log(t) / ( e^t - 1 ) dt
0

[Tiara]

Originally posted by CP5670
what was the original problem in that last quote?

Integrating Trig Function: dx/(3+4sin x)^2

Very simple I know...