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Integral on a derivative's answer won't give original function

Daqing Li
Posted 3 years ago 
Integrate[D[ArcSin[x],x],x]

The answer is

ArcTan[x/Sqrt[1-x^2]]

Is this a bug?
POSTED BY: Daqing Li
9 Replies
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To be precise, ArcTan[x/Sqrt[1-x^2]] is equivalent to ArcSin[x] when -1 < x < 1:

{ArcTan[x/Sqrt[1 - x^2]], ArcSin[x]} /. x -> 2.
ReImPlot[ArcTan[x/Sqrt[1 - x^2]] - ArcSin[x], {x, -5, 5}]

However, in both versions 12.0 and 13.2 from

Integrate[D[ArcSin[x], x], x]

I get ArcSin[x].
POSTED BY: Gianluca Gorni
Daqing Li
Daqing Li
Posted 3 years ago

Below is what I got

enter image description here
POSTED BY: Daqing Li
Daqing Li
Daqing Li
Posted 3 years ago

enter image description here
POSTED BY: Daqing Li
Eric Rimbey
Eric Rimbey
Posted 3 years ago

Try FullSimplify on that last expression
POSTED BY: Eric Rimbey
Eric Rimbey
Eric Rimbey
Posted 3 years ago
POSTED BY: Eric Rimbey

For me works fine. enter image description here

Regards M.I.
POSTED BY: Mariusz Iwaniuk
Daqing Li
Daqing Li
Posted 3 years ago

Yes, it is correct in math sense as ArcTan[x/Sqrt[1-x^2]] is the same as ArcSin[x]. However, it did caught me off guard when Integrate[D[ArcSin[x],x],x] did not give me ArcSin[x] directly.

POSTED BY: Daqing Li 
f = ArcSin[x];
D[Integrate[f, x], x] == f
(*True*)
POSTED BY: Mariusz Iwaniuk

Best way to check is to differentiate the integral, subtract the integrand, and see if the result is zero.
POSTED BY: Daniel Lichtblau
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