[Maxima-bugs] [maxima:bugs] #3691 Incorrect integral over asin(1-x^2) from -1 to 1

[Robert D. <rob...@us...>]

R. Toy reports the following to the Maxima mailing list:

Maxima computes `integrate(asin(1-x^2),x,-1,1)` using antideriv which
returns
```
x*atan((x^2-1)/(x*sqrt(2-x^2)))+2*sqrt(2-x^2)
```
To evaluate the definite integral, maxima uses intsubs and discovers
discontinuities (discontinuites-in-interval) at -1, 0, 1.  A plot of the
antiderivative does show a discontinuity at 0.  This means whole-intsubs
is used to compute the parts, and same-sheet-subs is used to evaluate
the parts.  For the interval [-1,0], it returns 2^(3/2)-1.  For [0, 1]
it returns 2-2^(3/2).  When these are summed up, we get 0.

I guess same-sheet-subs is thinking the sheets are different and returns
the negative of the expected value for [0,1].

Not that limit-subs gives up if the integrate involves atan.

However, I think the underlying problem is that the antiderivative is
not quite right.  asin is always positive in the [-1,1], so the
integral should be monotonic increasing.  But
```
x*atan((x^2-1)/(x*sqrt(2-x^2)))+2*sqrt(2-x^2) 
```
isn't. intsubs/same-sheet-subs is not choosing the correct branch of the atan
function.



---

** [bugs:#3691] Incorrect integral over asin(1-x^2) from -1 to 1**

**Status:** open
**Group:** None
**Labels:** integrate asin 
**Created:** Tue Dec 29, 2020 08:39 AM UTC by David Scherfgen
**Last Updated:** Tue Dec 29, 2020 08:39 AM UTC
**Owner:** nobody


This is incorrect:

~~~
(%i1)	integrate(asin(1-x^2),x,-1,1);
(%o1)	0
~~~

The correct result for the entire integral would be 2^(5/2)-4.
The integrand is symmetric.
Interestingly, integrating from 0 to 1 or from -1 to 0 gives the correct (half) result:

~~~
(%i2)	integrate(asin(1-x^2),x,0,1);
(%o2)	2^(3/2)-2
(%i3)	integrate(asin(1-x^2),x,-1,0);
(%o3)	2^(3/2)-2
~~~


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