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

Daqing Li
Posted 2 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 2 years ago

Below is what I got

enter image description here
POSTED BY: Daqing Li
Daqing Li
Daqing Li
Posted 2 years ago

enter image description here
POSTED BY: Daqing Li
Eric Rimbey
Eric Rimbey
Posted 2 years ago

Try FullSimplify on that last expression
POSTED BY: Eric Rimbey
Eric Rimbey
Eric Rimbey
Posted 2 years ago

I think this thread is trying to hint at a couple of things:

  • Integration and differentiation are not perfect inverses of each other. The most obvious reason is that you can always tack on a constant term to the anti-derivative, but there are other considerations, like being able to effectively ignore isolated points of discontinuity. So, some care might need to be taken if you're expecting the integral of the derivative to equal the original expression.
  • The expression ArcSin[x] is effectively equivalent to ArcTan[x/Sqrt[1 - x^2]], so in some sense you did get the original back.
POSTED BY: Eric Rimbey

For me works fine. enter image description here

Regards M.I.
POSTED BY: Mariusz Iwaniuk
Daqing Li
Daqing Li
Posted 2 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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