Integral of sec(x)

by cragar
Tags: integral, secx
 P: 2,466 $$sec(x) = \frac{2}{e^{ix}+e^{-ix}}$$ then i multply bot top and bottom by $$e^{ix}$$ so i can do a u substitution $$u=e^{ix} du=ie^{ix}$$ so then $$\int {\frac{2du}{(u^2+1)i}} =\frac {2arctan(u)}{i}}$$ so then i turn the arctan into a log then i get $$ln|e^{ix}+i|-ln|e^{ix}-i| + c$$ then how do i get the real part out if this .
 Sci Advisor HW Helper PF Gold P: 12,016 Well, you MIGHT do it that way, but a simpler integration would be to set: $$1=\cos^2(\frac{x}{2})+\sin^{2}\frac{x}{2}$$ $$\cos(x)=\cos^{2}(\frac{x}{2})-\sin^{2}(\frac{x}{2})$$ These identities implies: $$\sec(x)=\frac{1+\tan^{2}(\frac{x}{2})}{1-\tan^{2}(\frac{x}{2})}$$ Setting, therefore: $$u=\tan(\frac{x}{2})\to\frac{du}{dx}=\frac{1}{2}\sec^{2}(\frac{x}{2})=\f rac{1}{2}(1+u^{2})$$ You'll get a rational integrand in u that you can solve by partial fractions decomposition: $$\int\sec(x)dx=\int\frac{2du}{1-u^{2}}$$
 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 $$\frac{cos(x)}{1-(sin(x))^2}$$ then u=sin(x) and du=cos(x)
 HW Helper P: 3,348 Integral of sec(x) Combine the two log terms into one, and use $$Log z = \ln |z| + i Arg(z)$$. Ie the Real part is simply the natural log of the modulus.
 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 .
 HW Helper P: 3,348 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.