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Integral question - lrdheat - 08-12-2017 07:49 PM
Is there any way to get ((pi)^2)/4 for the integral from 0 to pi of (x*(sin x))/(1+ (cos x)^2) ?
RE: Integral question - Helge Gabert - 08-13-2017 12:15 AM
Yes, it can be done.
Note that arctan(1)-arctan(0) is being recognized as the desired symbolic solution (pi^2/4).
Just kidding. The hard part is to get there through some clever symmetrical substitution like u=pi-x, and some other substitutions, a shown here
https://artofproblemsolving.com/community/c7h542808_integrate_xsinx1cos2_x__on_0_pi
Not sure if that recognition pattern has been implemented in Giac/Xcas (maybe it is too expensive).
RE: Integral question - Mark Hardman - 08-13-2017 01:00 AM
(08-12-2017 07:49 PM)lrdheat Wrote: Is there any way to get ((pi)^2)/4 for the integral from 0 to pi of (x*(sin x))/(1+ (cos x)^2) ?
ibpu((x*sin(x)/(1+cos(x)^2)),x,x,0,π)
RE: Integral question - Helge Gabert - 08-13-2017 01:24 AM
Excellent! Didn't think about ibpu and ibpdv. That'll do it.
RE: Integral question - lrdheat - 08-13-2017 03:24 PM
Thanks!