I am calculating the arc length of a piece of parabola. Here is the
integral,
'integrate(sqrt(9*x^2+37),x,0,2);
I would like Maxima to do the problem using the following
substitution:
changevar(%,3*x=sqrt(37)*sinh(t),t,x);
Maxima returns,
37*('integrate(cosh(t)*sqrt(sinh(t)^2+1),t,0,asinh(6/sqrt(37))))/3
%,nouns;
`rat' replaced .8717198386293096 by 8936/10251 = .8717198322114915
`rat' replaced .8717198386293096 by 8936/10251 = .8717198322114915
`rat' replaced .8717198386293096 by 8936/10251 = .8717198322114915
`rat' replaced .8717198386293096 by 8936/10251 = .8717198322114915
`rat' replaced .8717198386293096 by 8936/10251 = .8717198322114915
`rat' replaced .8717198386293096 by 8936/10251 = .8717198322114915
`rat' replaced .8717198386293096 by 8936/10251 = .8717198322114915
`rat' replaced 1.128616979745806 by 9205/8156 = 1.128616969102501
(%o402) 340585/24468
This is wrong. Yet, the original integral is done correctly, i.e.,
(%i403) %o398,nouns;
(%o403) (37*asinh(6*sqrt(37)/37)+6*sqrt(73))/6
Also, if I do the indefinite integral obtained after the substitution, i.e.,
integrate(cosh(t)^2,t);
and the substitute the limits (and multiply with the constants),
everything is fine.
Similar thing happened earlier today working on the arc of parabola
y=4-x^2 (in the first quadrant). This is on Maxima 5.13, Clisp 2.41.
I do not remember a similar problem being discussed, nor did I find in
the bug database.
Milan
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