Bill Wood wrote:
> On Tue, 2007-05-15 at 20:23 -0500, Doug Stewart wrote:
>
>> I was showing my students how to use Maxima to check their homework.
>>
>> When I got to this part of the example I got a surprise.
>> integrate(sin(x)*cos(x),x)
>>
>> maxima said it was -cos^2(x)/2 and my book said it was sin^2(x)/2 so I
>> looked it up in shaums formula book
>> and it said sin^2(x)/2.
>>
>> So I then plotted up a graph in Octave and was convinced that Maxima is
>> wrong.
>>
>> I then submitted a bug report and someone replied and told me that I
>>
>
> You don't even need calculus -- sin^2(x) + cos^2(x) = 1 is a trig
> identity. That still doesn't justify rudeness. The fact that this is a
> well-known identity does not imply that someone will see how to apply it
> in just the right form in a particular situation.
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>
Yes you are all correct. :-) I see now what happened.
I have been doing all the definite integrals for different Fourier
series expansions, and was thinking that in stead of doing the definite
integral complete, I would show them how to use Maxima to get all the
parts before it was evaluated at 0 and pi.
etc.
So my mind was totally working with definite integrals and I forgot that
I was now asking for an indefinite integral from Maxima.
It worked for all the other examples that I had shown them.
It would be nice if this one would return sin^2(x)/2 instead of -cos^(x)/2.
(yes I know that they only differ by a constant and yes I know
sin^2+cos^2=1 etc.)
Thanks for your help.
Doug Stewart