I need help with the following:
$\displaystyle \int \frac{1}{\cos{x}+i\sin{x}} dx$ = ?
Thanks
$\displaystyle \int \frac{1}{cosx + isinx}~dx$
$\displaystyle = \int \frac{1}{e^{ix}}~dx$
$\displaystyle \left(e^{ix} = cosx + isinx\right)$ <--- Euler's Formula
$\displaystyle =\int e^{-ix} ~dx$
$\displaystyle =\int cosx - isinx ~dx$
$\displaystyle \left(e^{-ix} = e^{i(-x)} = cos(-x) + isin(-x) = cosx - isinx\right)$ <--- Euler's Formula for -x
$\displaystyle = sinx + icosx$
But wait, so you guys are saying that
$\displaystyle \int \frac{1}{e^{ix}}~dx=\int e^{ix}~dx$ ?
and that, since the antiderivative of e^x is e^x,
$\displaystyle \frac{1}{e^{ix}}= e^{ix}$
?
That would contradict $\displaystyle \ln{x}=\int_{1}^{x}\frac{1}{x}~dx$
wouldn't it?
Okay sorry. I spazzed out there for a second. Now I got it.
But while we're on the subject, the wikipedia page on i does say that i= -i.
If that is true, then wouldn't it follow that:
$\displaystyle \frac{1}{e^{ix}}=e^{ix}$ ?
The wikipedia quote Verbatim:
"If all mathematical textbooks and published literature referring to imaginary or complex numbers were rewritten with −i replacing every occurrence of +i (and therefore every occurrence of −i replaced by −(−i) = +i), all facts and theorems would continue to be equivalently valid"
Basically the concept is that $\displaystyle i$ arose from solving $\displaystyle x^2+1=0$. But the FTA tells us that there must be two solutions, obviously $\displaystyle x=i,-i$. As a convention then we take $\displaystyle \sqrt{-1}=i$ (sort of). We could have, with equal validity, make the convention that $\displaystyle \sqrt{-1}=-i$ in which case everything would be different. But this is no way means that $\displaystyle i=-i$