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I know that $\arcsin(\sin(x))$ would be $x$ (in the right domain). But, I'm unsure how to handle the scalar multiplication. Any thoughts on how to proceed?

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    $\begingroup$ There is no "nicer" expression for this. $\endgroup$ Commented Feb 6, 2021 at 8:26
  • $\begingroup$ Why do you need such "simplification"? What are you trying to integrate? $\endgroup$
    – Robert Z
    Commented Feb 6, 2021 at 8:27
  • $\begingroup$ what an answer should give you? $\endgroup$ Commented Feb 6, 2021 at 8:34
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    $\begingroup$ You could write it as $-i\ln\left(e^{ix}-e^{-ix}+\sqrt{e^{2ix}-1+e^{-2ix}}\right)$, but to me that looks worse. In fact, it also hints at why we can't do better than $\arcsin(2\sin x)$. $\endgroup$
    – J.G.
    Commented Feb 6, 2021 at 13:09

1 Answer 1

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You can write $$\arcsin(a\sin (x))=\sum_{n=0}^\infty \frac{ (2 n)!\,\, a^{2 n+1} }{4^n\,(2 n+1)\, (n!)^2}\,\sin ^{2 n+1}(x)$$

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  • $\begingroup$ Hi, but do you remember all the formulas with your mind? 😊 +1. $\endgroup$
    – Sebastiano
    Commented Feb 6, 2021 at 9:56
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    $\begingroup$ @Sebastiano. At least I try (hard !). I felt in love with infinite series $64+$ years ago. In fact, take the infinite series of $\arcsin(t)$ and make $t=a\sin(x)$. Simple, no ? Cheers :-) $\endgroup$ Commented Feb 6, 2021 at 10:00
  • $\begingroup$ Ah...there was a reason 😄😄....my sincere compliments. $\endgroup$
    – Sebastiano
    Commented Feb 6, 2021 at 10:01
  • $\begingroup$ @Sebastiano. Compliments for what ? My age ? My memory ? Back to serious, thanks :-) $\endgroup$ Commented Feb 6, 2021 at 10:09
  • $\begingroup$ For the answers and....for the memory 😊 $\endgroup$
    – Sebastiano
    Commented Feb 6, 2021 at 10:10

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