S Sin^6x dx
= S Sin^6xcosx / cosx dx ( * by cosx/cosx)
let u = sinx, du = cosx dx
= S u^6 / cosx du but u = sinx, therefoce sqr(1-u^2) = cosx
= S u^6 / sqr(1-u^2) du - which is simple to integrate
hehe usually i dont post anymore but im in a good mood from my umat scores :)
(btw it would be good if someone got latex or some math program for this board like other maths forums do)
anyway
area under y = lnx (from 1 to n) > area under sum of trapeziums from 1 to n
nlnn - n + 1 >...