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Thread: antiderivative involving euler's formula

  1. #1
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    antiderivative involving euler's formula

    I need help with the following:

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



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    Super Member Aryth's Avatar
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    $\displaystyle \int \frac{1}{cosx + isinx}~dx$

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

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

    $\displaystyle = sinx + icosx$
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by rainer View Post
    I need help with the following:

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



    Thanks
    You don't exactly need Euler's formula here. Multiply by the algebraic conjugate of denominator and watch the problem melt away.
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    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?
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    Quote Originally Posted by rainer View Post
    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?
    Nothing in either post implies $\displaystyle \int \frac{1}{e^{ix}}~dx=\int e^{ix}~dx$. Both ways give the result of

    $\displaystyle \int \frac{1}{e^{ix}}~dx = sinx + icosx \neq cosx + isinx = e^{ix}$
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    Quote Originally Posted by Defunkt View Post
    Nothing in either post implies $\displaystyle \int \frac{1}{e^{ix}}~dx=\int e^{ix}~dx$. Both ways give the result of

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

    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:

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


    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"
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by rainer View Post
    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"
    $\displaystyle i=-i=i^3\implies i^2=1$. Sweet!

    Can you find the subtle difference in what was meant to be said and what was said?
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    Say what?
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by rainer View Post
    Say what?
    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 $\displaystyle i=-i$
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