[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 .
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})=\frac{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]
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)
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.
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 .
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.
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. =]