Wzory rekurencyjne dx In = n " N Niech , +" n (1+ x2) I1 = arctgx + C Wtedy ; n e" 2 a dla : 2 u = x u =1 1+ x2 - x2 2x Å" x -n+1 1 In = dx = In-1 - dx = -n (1+ x2) = 2 +" n +" n 2 v = 2x(1+ x2) v = (1+ x2) (1+ x2) - n +1 ëÅ‚ öÅ‚ x 1 x 1 1 ìÅ‚ = In-1 - - dx÷Å‚ = In-1 + - In-1 = 2 n-1 +" n-1 n-1 ìÅ‚ ÷Å‚ 2n - 2 (1- n)(1+ x2) (1- n)(1+ x2) (2n - 2)(1+ x2) íÅ‚ Å‚Å‚ 2n - 3 x = In-1 + n-1 2n - 2 (2n - 2)(1+ x2) Podobnie n n-2 1 n-1 cos x Å" sinn-1 x + dla +"sin xdx = - n n +"sin xdx n e" 2 n n-2 1 n-1 x dla n n +"cos xdx = sin x Å" cosn-1 + +"cos xdx n e" 2 n = 2 dla mamy: 1 1 2 +"sin xdx = 2 x - 2 cos xsin x + C 1 1 2 +"cos xdx = 2 x + 2 sin x cos x + C opracowaÅ‚ PaweÅ‚ Sztur 1