Integral question - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: HP Prime (/forum-5.html) +--- Thread: Integral question (/thread-8827.html) |
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! |