# Thread: antiderivative involving euler's formula

1. ## antiderivative involving euler's formula

I need help with the following:

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

Thanks

2. $\int \frac{1}{cosx + isinx}~dx$

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

$\left(e^{ix} = cosx + isinx\right)$ <--- Euler's Formula

$=\int e^{-ix} ~dx$

$=\int cosx - isinx ~dx$

$\left(e^{-ix} = e^{i(-x)} = cos(-x) + isin(-x) = cosx - isinx\right)$ <--- Euler's Formula for -x

$= sinx + icosx$

3. Originally Posted by rainer
I need help with the following:

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

Thanks
You don't exactly need Euler's formula here. Multiply by the algebraic conjugate of denominator and watch the problem melt away.

4. But wait, so you guys are saying that

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

and that, since the antiderivative of e^x is e^x,

$\frac{1}{e^{ix}}= e^{ix}$

?

That would contradict $\ln{x}=\int_{1}^{x}\frac{1}{x}~dx$

wouldn't it?

5. Originally Posted by rainer
But wait, so you guys are saying that

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

and that, since the antiderivative of e^x is e^x,

$\frac{1}{e^{ix}}= e^{ix}$

?

That would contradict $\ln{x}=\int_{1}^{x}\frac{1}{x}~dx$

wouldn't it?
Nothing in either post implies $\int \frac{1}{e^{ix}}~dx=\int e^{ix}~dx$. Both ways give the result of

$\int \frac{1}{e^{ix}}~dx = sinx + icosx \neq cosx + isinx = e^{ix}$

6. Originally Posted by Defunkt
Nothing in either post implies $\int \frac{1}{e^{ix}}~dx=\int e^{ix}~dx$. Both ways give the result of

$\int \frac{1}{e^{ix}}~dx = sinx + icosx \neq cosx + isinx = e^{ix}$

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:

$\frac{1}{e^{ix}}=e^{ix}$ ?

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"

7. Originally Posted by rainer
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:

$\frac{1}{e^{ix}}=e^{ix}$ ?

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"
$i=-i=i^3\implies i^2=1$. Sweet!

Can you find the subtle difference in what was meant to be said and what was said?

8. Say what?

9. Originally Posted by rainer
Say what?
Basically the concept is that $i$ arose from solving $x^2+1=0$. But the FTA tells us that there must be two solutions, obviously $x=i,-i$. As a convention then we take $\sqrt{-1}=i$ (sort of). We could have, with equal validity, make the convention that $\sqrt{-1}=-i$ in which case everything would be different. But this is no way means that $i=-i$