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Integral of Sqrt[1-x^2] appears wrong in Wolfram Alpha

Posted 2 days ago

Hello. I asked Wolfram Alpha the question shown below. Shown below is exactly what I typed into the Wolfram Alpha query field. (I also typed into the query field variations of the text below. I got the same result.)

Integrate[Sqrt[1- x^2], x]

Here's the answer it gave me. (For each expression below, the format is Wolfram language plain text.)

(x Sqrt[1- x^2])/2- ArcTan[Sqrt[1- x^2]/(1+ x)]

IMO the above answer is wrong.

I used Mathematica and here's the answer it gave me:

(x Sqrt[1- x^2]+ ArcSin[x])/2

IMO the answer directly above is the correct one.

I also solved the integral by hand, using trig substitution (it's a well-known integral), and I got the same answer that Mathematica gave me.

Would someone please explain. Or at least, would someone please ask Wolfram Alpha the same question shown above, and tell us the result. IMO the result should not contain the ArcTan function. It should contain the ArcSin function.

POSTED BY: M Bennett
2 Replies

Antiderivatives may differ by a constant over connected components of the domain. There is not just one antiderivative that is "right." Trying plotting the difference of the two expressions. It should consist of horizontal lines, perhaps more than one since Sqrt[expr] may have branch cuts. Sometimes one may need to plot something of the form ReImPlot[ (* difference *), {x, a, b}] or ComplexPlot3D[(* difference *), {x, a1 + a2*I, b1 + b2*I}], since Integrate[] really deals with antiderivatives over the complex plane where the branch cuts live.

Also consider that trig identities mean some functions may be written in terms of other functions.

FullSimplify[ArcTan[x/Sqrt[1 - x^2]] == ArcSin[x], -1 < x < 1]
(*  True  *)

So ArcTan[] may be expressed in terms of ArcSin[] and vice versa.

POSTED BY: Michael Rogers
Posted 1 day ago

Thank you for the nice explanation.

I see that you used the FullSimplify function and specified a certain range for x; then tested whether the 2 functions ArcTan and ArcSin are equal over that range. We see that the result is True.

Before I posted the question, I used Mathematica to test whether the functions ArcTan and ArcSin are equal, but I had not used the FullSimiplify function, or specified a range.

POSTED BY: M Bennett
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