I need help with the following:

$\displaystyle \int \frac{1}{\cos{x}+i\sin{x}} dx$ = ?

Thanks

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- Dec 1st 2009, 09:50 AMrainerantiderivative involving euler's formula
I need help with the following:

$\displaystyle \int \frac{1}{\cos{x}+i\sin{x}} dx$ = ?

Thanks - Dec 1st 2009, 12:47 PMAryth
$\displaystyle \int \frac{1}{cosx + isinx}~dx$

$\displaystyle = \int \frac{1}{e^{ix}}~dx$

$\displaystyle \left(e^{ix} = cosx + isinx\right)$ <--- 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

$\displaystyle = sinx + icosx$ - Dec 1st 2009, 01:57 PMDrexel28
- Dec 1st 2009, 03:38 PMrainer
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? - Dec 1st 2009, 03:47 PMDefunkt
- Dec 2nd 2009, 06:40 AMrainer

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" - Dec 2nd 2009, 06:41 AMDrexel28
- Dec 2nd 2009, 06:54 AMrainer
Say what?

- Dec 2nd 2009, 06:57 AMDrexel28
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 $\displaystyle i=-i$