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Is there any way to get ((pi)^2)/4 for the integral from 0 to pi of (x*(sin x))/(1+ (cos x)^2) ?

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/communit...x__on_0_pi

Not sure if that recognition pattern has been implemented in Giac/Xcas (maybe it is too expensive).

(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)

Excellent! Didn't think about ibpu and ibpdv. That'll do it.

Thanks!