## integral of sec(x)

$$sec(x) = \frac{2}{e^{ix}+e^{-ix}}$$
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 .
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 Recognitions: Gold Member Homework Help Science Advisor 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: $$\int\sec(x)dx=\int\frac{2du}{1-u^{2}}$$
 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)

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## 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.
 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 .
 Recognitions: Homework Help 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.

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