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Definite integrals

  1. Nov 8, 2004 #1
    can anyone help me with this problem...
    the integral from 0 to half pi: 1/(1+cosx)dx...
    Thanks...
     
  2. jcsd
  3. Nov 8, 2004 #2

    arildno

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    Hint use the relation:
    [tex]\cos^{2}(\frac{x}{2})=\frac{1+\cos(x)}{2}[/tex]
     
  4. Nov 8, 2004 #3
    from my perspective, that problem would be very hard to solve without changing it somehow. try multiplying the top and bottom by (1-cos x) and see what happens. remember that sin^2 x + cos^2 x = 1.

    that is the method i used, but it did require me to use the reduction formula on one of the integrals. hopefully that helps, thought it may not be the inteded method.
     
  5. Nov 9, 2004 #4
    so 1/(1+cosx)=(1-cosx)/(1-cos^2x)=(1-cosx)/sin^2x=(1/sin^2x)-(cosx/sin^2x)...
    Aha~now i get it
    the integration of 1/sin^2x is -cotx, and cosx/sin^2x=cosx*sin^(-2)x, and the integral of this is -1/sinx
    so the integration of 1/(1+cosx) is -cotx+1/sinx, but now i have a problem, cot(half pi) does not exist...
    help still needed...
     
  6. Nov 9, 2004 #5
    i forgot that the derivative of cot is -csc^2. you don't need the reduction formula after all! i worked this out and i got

    1/sin x - cot x
    or
    1/sin x - cos x/sin x

    the problem is when it approaches zero, not pi/2. to me it looks like this is divergent, thus there is no definant area (as it approaches zero from the right, 1/sin x approaches infinity).

    maybe i did something wrong, but the math looks correct. as a forewarning i'm just finishing the last chapter in calc 2 in college, so i'm not a math guru.

    edit: i don't remember if limit laws dictate that you could say infinity minus infinity equals a total of zero. that could be it's been almost a year since i took calc 1. if so, you'd have (1-0)-(infinity - infinity = 0) = 1 unit^2.

    i don't know when i get problems wrong on tests its usually due to magic algebra, that could be a case of it right there. i'd look it up but i need to finish writing up my lab for this evening.
     
    Last edited: Nov 9, 2004
  7. Nov 10, 2004 #6
    The answer is 1. Or maybe we should convert (1/sin^2x)-(cosx/sin^2x) so that we can substitute half pi and 0 in it...not sure...
     
  8. Nov 10, 2004 #7
    well, sin of 0 is 0, so either way you're dividing both by zero. the only thing i can figure out is that you do it as an improper integral and the infinities cancel eachother.
     
  9. Nov 10, 2004 #8
    I suggest you use the advice of arildno. It will be the best way to solve this.

    [tex]\int\frac{2}{\cos(x)+1}*dx = \int\frac{1}{\cos^{2}(\frac{x}{2})} * dx[/tex]

    and you know that [tex]dx = 2*d(\frac{x}{2})[/tex]
    and
    [tex]\int\frac{1}{\cos^{2}(x)} = \tan(x)[/tex]

    regards
    marlon
     
    Last edited: Nov 10, 2004
  10. Nov 10, 2004 #9
    that is clever, i didn't remember that identity. but you aren't finished. remember that the integral is from 0 to pi/2. under that integral the function is divergent, thus the area doesn't exist. right? (lim t---> pi/2 [tan x] from 0 to t) sin x /cos x -- as x approaches pi/2, thus cos x = 0, bad things happen.

    also divergent by relation to 1/cos x isn't it?

    if anyone has a certain answer, please elucidate! (no mathmatica at home :( )
     
  11. Nov 10, 2004 #10
    edit: nevermind, i thought you were saying the answer was tan x. disregard this and above message!
     
    Last edited: Nov 10, 2004
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