integrate(1/(sin(x)^2+1),x,a,b)?



Ray,

Sorry that wasn't clear. The Maple 10 result is a split-rule function:
For z < -Pi / 2, the value is
(arctan(tan(z)*2^(1/2))+ceil(1/2*(-Pi+2*z)/Pi)*Pi) / sqrt(2), for
z = -Pi / 2, the value is -Pi sqrt(2) / 4, and ....

But outside the interval [-%pi/2, %pi/2] the Maple result is wrong.
Maybe this is controlled by a switch. I don't know much about Maple.

I think a better (than Maple 10 or Macsyma) antiderivative is

  atan2(sqrt(2) * sin(x), cos(x)) / sqrt(2) + sqrt(2) * %pi *
  floor((x-%pi)/(2*%pi));

It's graph looks continuous and

(%i14) atan2(sqrt(2) * sin(x), cos(x)) / sqrt(2) + sqrt(2) * %pi *
floor((x-%pi)/(2*%pi))$

(%i15) trigsimp(diff(%,x));
(%o15)
((sqrt(2)*%pi*sin(x)^2+sqrt(2)*%pi)*('diff(floor((x-%pi)/(2*%pi)),x,1))+1)/(sin(x)^2+1)


With 'diff(floor(...)) --> 0, this also looks OK.  The atan2 function is
pretty
handy for antiderivatives.

Barton

-----maxima-bounces at math.utexas.edu wrote: -----

>To: Barton Willis
>From: Raymond Toy
>Sent by: maxima-bounces at math.utexas.edu
>Date: 06/27/2007 02:29PM
>cc: Maxima List
>Subject: Re: [Maxima] integrate(1/(sin(x)^2+1),x,a,b)?
>
>>>>>> "Barton" == Barton Willis  writes:
>
>    Barton> Here is what I get using Maple 10
>[snip]
>    >> integrate(1/(sin(x)^2+1),x);
>    >>
>
>    Barton>                           1/2                1/2
>    Barton>                      1/2 2    arctan(tan(x) 2   )
>
>    >> _EnvAllSolutions := true:
>    >> integrate(1/(sin(x)^2+1),x=0..z);
>    >>
>
>    Barton>   {
>    Barton>   {      1/2 /               1/2         -Pi + 2 z    \
>  Pi
>    Barton>   { 1/2 2    |arctan(tan(z) 2   ) + ceil(---------) Pi| , z <
>- ----
>    Barton>   {          \                             2 Pi       /
>  2
>
>    Barton>               1/2
>    Barton>           Pi 2             Pi
>    Barton>         - ------- , z = - ----
>    Barton>              4             2
>
>    Barton>                            1/2   1/2        Pi
>    Barton>         1/2 arctan(tan(z) 2   ) 2    , z < ----
>    Barton>                                             2
>
>    Barton>             1/2
>    Barton>         Pi 2           Pi
>    Barton>         ------- , z = ----
>    Barton>            4           2
>
>    Barton>              1/2 /               1/2             -Pi + 2 z
> \
>    Barton>         1/2 2    |arctan(tan(z) 2   ) + Pi floor(---------) +
>Pi| ,
>    Barton>                  \                                 2 Pi
> /
>
>    Barton>          Pi
>    Barton>         ---- < z
>    Barton>          2
>
>
>I'm sorry, I can't understand what this is trying to tell me.  I see
>there are 3 possible answers?  But what are those z things?
>
>Ray
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