[Maxima-bugs] [ maxima-Bugs-2651868 ] integrate(sin(a*x+b)^2,x); incorrect answer.

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Bugs item #2651868, was opened at 2009-03-01 16:12
Message generated for change (Comment added) made by digital7
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Category: Lisp Core - Integration
Group: None
>Status: Open
Resolution: Invalid
Priority: 5
Private: No
Submitted By: Denis (digital7)
Assigned to: Nobody/Anonymous (nobody)
Summary: integrate(sin(a*x+b)^2,x); incorrect answer.

Initial Comment:
(%i39) integrate(sin(a*x+b)^2, x);
(%o39) (-sin(2*(a*x+b))/2+a*x+b)/(2*a)

And after that:

(%i42) expand((-sin(2*(a*x+b))/2+a*x+b)/(2*a));
(%o42) -sin(2*a*x+2*b)/(4*a)+x/2+b/(2*a)

But using risch function and integrating myself I've got another results:

(%i40) risch(sin(a*x+b)^2, x);
(%o40) -(sin(2*a*x+2*b)-2*a*x)/(4*a)


sin(x)^2 = 1/2 * (1 - cos(2*x)) =>

sin(a*x+b)^2 = 1/2 * (1 - cos(2*a*x + 2*b))

\                         \     \
|sin(a*x+b)^2 dx = 1/2 * (|dx - |cos(2*a*x + 2*b)dx) =
\                         \     \

= x/2 - 1/(2*a) * sin(2*a*x + 2*b)

What does "b/(2*a)" mean? Where have maxima got it?

P.S. I've attached a wxMaxima session file with these calculations.

Best regards,
Denis.

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>Comment By: Denis (digital7)
Date: 2009-03-01 19:24

Message:
Thank you for your answer. Because I was calculating a definite integral
for my complex work and checking the integration if Maxima, I haven't
noticed that b/(2*a) is a constant.

Some time later I'll dig into the integration routines. :)

I think that topic is closed. Thank you again.

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Comment By: Raymond Toy (rtoy)
Date: 2009-03-01 18:00

Message:
Both results are correct.  They only differ by the constant b/(2*a).  The
derivative of each, after some manipulations, equals the integrand.

Not sure what you're asking about "b/(2*a)".   If you really want to know
how maxima got it you'll have to dig into the integration routines.

Marking as pending/invalid.   Please update if this analysis is incorrect.

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