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



Oh...my antiderivative was wrong --- something like

  atan2(sqrt(2)*sin(x),cos(x))/sqrt(2)+sqrt(2)*%pi*ceiling((x-%pi)/(2*%pi))
  + %pi * sqrt(2) for x > 0
  atan2(sqrt(2)*sin(x),cos(x))/sqrt(2)+sqrt(2)*%pi*floor((x-%pi)/(2*%pi)),
  for x < 0

might be correct . Maybe this can be cunched to a non-split rule...(round
instead of floor / ceiling?)

Barton

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

>To: Raymond Toy <raymond.toy at ericsson.com>
>From: Barton Willis <willisb at unk.edu>
>Sent by: maxima-bounces at math.utexas.edu
>Date: 06/27/2007 03:10PM
>cc: Maxima List <maxima at math.utexas.edu>
>Subject: Re: [Maxima] 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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