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Help with this integral

  1. Nov 26, 2004 #1
    I have some problems in calculating the integral dx/cos(x) i have made the change exp(ix)=t and get the solution (2/i)artan(exp(ix)) but what is the real number solution?.thanks
     
  2. jcsd
  3. Nov 26, 2004 #2

    arildno

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    It's better to use the half-angle formulas:
    [tex]\cos^{2}(\frac{x}{2})-\sin^{2}(\frac{x}{2})=\cos(x)[/tex]
    [tex]\cos^{2}(\frac{x}{2})+\sin^{2}(\frac{x}{2})=1[/tex]
    Your integral is then easy to evaluate with respect to the variable:
    [tex]u=tan(\frac{x}{2})[/tex]
     
  4. Nov 26, 2004 #3

    Maybe I am missing something, but isn't dx/cos(x) = sec(x)dx. Doing that integral is very easy, you should be able to handle it.
     
  5. Nov 27, 2004 #4
    1/cos(x)dx
    =cos(x)/(1-sin^2(x))dx
    =1/(1-sin^2(x))d(sin(x))
     
  6. Nov 28, 2004 #5
    To SomeRandomGuy,

    Do you tutor? I have a kid who could use some help with his math homework. No doubt he could benefit from your penetrating insight and clarity of exposition.
     
  7. Nov 28, 2004 #6
    :rofl:

    Here's yet another method to do this integral:

    [tex]\int \sec x \; dx = \int \sec x \; \frac{\sec x+\tan x}{\sec x+\tan x} \; dx[/tex]

    [tex]\int \frac{\sec^{2}x + \sec x\tan x}{\sec x+\tan x}\;dx[/tex]

    Take [itex]u=\sec x+\tan x[/itex], then [itex]\frac{du}{dx}=\sec x\tan x+\sec^{2}x[/itex].

    [tex]\int \frac{\sec^{2}x + \sec x\tan x}{\sec x+\tan x} \; dx = \int \frac{1}{u} \; du = \ln(u) \; + \; C = \ln(\sec x+\tan x) \; + \; C[/tex]
     
  8. Nov 28, 2004 #7
    I am one to appreciate sarcasm as I am a very sarcastic individual myself. I am a tutor, actually, and I will rarely complete an integral or a problem for someone I tutor. Instead, I try to have them figure it out for themselves. That way, it is more likely they will remember the process and what to look for.
     
  9. Nov 29, 2004 #8

    dextercioby

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    Why complicate ?
    [tex] \int \frac{dx}{\cos x}=\int \frac{d(\sin x)}{1-\sin^2 x}=\frac{1}{2}[\int \frac{d(\sin x)}
    {1-\sin x} +\int \frac{d(\sin x)}{1+\sin x}] =...=\frac{1}{2} \ln(\frac{1+\sin x}{1-\sin x})+C
    =...=\ln [\frac{\tan(\frac{x}{2})+1}{\tan(\frac{x}{2})-1}]+C.[/tex]
     
  10. Nov 29, 2004 #9
    dextercioby, your method was already posted. I was only suggesting another method, which I personally don't think is complicated.
     
  11. Nov 29, 2004 #10

    dextercioby

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    Maybe it was posted...
    Check out that horrible trick u pulled in order to integrate secant of x.You wouldn't expect someone to think of it as simple...Obvious...Besides,it requires a lotta derivatives and it uses the secant function which is (as stated by me) unnecessary...
    It don't matter really how u pull an integral off.The trick is to do it as elegantly as possible.As comprehendable as possible.I assume u're not a teacher/never be one. :wink:
     
  12. Nov 29, 2004 #11
    Erm.. I wasn't trying to do the integral "as elegantly as possible". I was merely providing another another way to do it, in addition to the two methods already posted...
     
  13. Nov 29, 2004 #12

    dextercioby

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    Well,Mr.Devious,apparently u challanged me:
    [tex] \begin {array} {cc} \int \frac{dx}{\cos x} = \int \sec x dx = \int \sec x (\frac {\tan x}{\tan x })dx
    =^{d(\sec x)=\sec x\tan x dx} \int \frac{d(\sec x)}{\sin x \sec x}
    \\=^{\sin x=\sqrt{1-\frac{1}{\sec^{2} x}}} \int \frac{d(\sec x)}{\sqrt{\sec^{2} x-1}}
    =^{\sec x= \cosh \lambda ; \lambda=\cosh^{-1}(\sec x)}
    \int \frac{(\sinh \lambda) d\lambda}{\sinh \lambda}
    \\=\lambda +C= \cosh^{-1}(\sec x) +C = \ln(\sec x+\tan x)+C \end{array}
    [/tex]
    ,where,of course,the expression [tex] f^{-1}(x) [/tex] strands for the inverse of the function [tex] f(x) [/tex].It stood for "argcosh (sec x)",but i couldn't find the function,so that's why i used this notation.

    Daniel.
     
    Last edited: Nov 29, 2004
  14. Nov 29, 2004 #13
    Challenged you? Why are you taking this personally? :rofl:

    [tex]\int \sec x \; dx = \int \frac{\cos x}{\cos^{2} x} \; dx = \int \frac{\cos x}{1-\sin^{2} x} \; dx [/tex]

    Take [itex]u=\sin x[/itex], then [itex]du=\cos x\;dx[/itex]. So:

    [tex]\int \sec x \; dx = \int \frac{1}{1-u^{2}} \; du = \tanh^{-1} u \; + \; C = \tanh^{-1}(\sin x) \; + \; C[/tex]

    :tongue2:
     
    Last edited: Nov 29, 2004
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