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Integral of sec(x)

by cragar
Tags: integral, secx
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cragar
#1
Jun13-10, 04:49 AM
P: 2,466
[tex]
sec(x) = \frac{2}{e^{ix}+e^{-ix}}
[/tex]
then i multply bot top and bottom by [tex] e^{ix} [/tex]
so i can do a u substitution
[tex] u=e^{ix} du=ie^{ix} [/tex]
so then [tex] \int {\frac{2du}{(u^2+1)i}}
=\frac {2arctan(u)}{i}} [/tex]
so then i turn the arctan into a log
then i get [tex] ln|e^{ix}+i|-ln|e^{ix}-i| + c [/tex]
then how do i get the real part out if this .
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arildno
#2
Jun13-10, 05:28 AM
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Well, you MIGHT do it that way, but a simpler integration would be to set:
[tex]1=\cos^2(\frac{x}{2})+\sin^{2}\frac{x}{2}[/tex]
[tex]\cos(x)=\cos^{2}(\frac{x}{2})-\sin^{2}(\frac{x}{2})[/tex]
These identities implies:
[tex]\sec(x)=\frac{1+\tan^{2}(\frac{x}{2})}{1-\tan^{2}(\frac{x}{2})}[/tex]
Setting, therefore:
[tex]u=\tan(\frac{x}{2})\to\frac{du}{dx}=\frac{1}{2}\sec^{2}(\frac{x}{2})=\f rac{1}{2}(1+u^{2})[/tex]

You'll get a rational integrand in u that you can solve by partial fractions decomposition:
[tex]\int\sec(x)dx=\int\frac{2du}{1-u^{2}}[/tex]
cragar
#3
Jun13-10, 06:38 AM
P: 2,466
sorry i should have said i want to see it done with complex numbers ,
I have done it that way before . but i wrote it like
[tex] \frac{cos(x)}{1-(sin(x))^2} [/tex]
then u=sin(x) and du=cos(x)

Gib Z
#4
Jun13-10, 06:49 AM
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Integral of sec(x)

Combine the two log terms into one, and use [tex]Log z = \ln |z| + i Arg(z)[/tex]. Ie the Real part is simply the natural log of the modulus.
cragar
#5
Jun13-10, 07:22 AM
P: 2,466
thanks for all of your answers guys , im not sure what modulus is i tired looking it up
could you maybe tell me where to read about it i have only had calc 3 .
Gib Z
#6
Jun13-10, 07:30 AM
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I'm sure you have if your doing integration like this! The modulus of a complex number a+bi is sqrt(a^2+b^2). You can think of it as the length of the line that connects the origin to a+bi on the Argand Plane.
arildno
#7
Jun13-10, 08:45 AM
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Quote Quote by Gib Z View Post
I'm sure you have if your doing integration like this! The modulus of a complex number a+bi is sqrt(a^2+b^2). You can think of it as the length of the line that connects the origin to a+bi on the Argand Plane.
Wessel Plane, if I may.
http://en.wikipedia.org/wiki/Caspar_Wessel
Gib Z
#8
Jun13-10, 08:59 AM
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Ahh my mistake !

In mathematics often things aren't named after who really should have gotten credit for them! There's a joke that for an entire century after Euler, to ensure other mathematicians got some recognition, things were named after the first person after Euler to discover it. =]


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