∫sin^n(x)dx
let Ln = ∫sin^nxdx
also
Ln =∫sin^(n - 1)x.sinxdx
from...........d(cosx) = -sinx dx
Ln = -∫sin^(n - 1)x d(cosx)
using integration by parts
dv = d(cosx)............ v = cosx.......................... u = sin^(n - 1)x............... du/dx = (n - 1)sin^(n - 2)x.cosx
Ln = -{cosx.sin^(n - 1) - (n - 1)∫cosx.sin^(n - 2)x.cosx dx}
= -cosx.sin^(n - 1) + (n - 1)∫cos^2(x).sin^(n - 2)x dx
= -cosx.sin^(n - 1) + (n - 1)∫sin^(n -2)x{1 - sin^2(x)} dx
= -cosx.sin^(n - 1) + (n - 1)∫sin^(n - 2)x dx - (n - 1)∫sin^nx dx
Ln = -cosx.sin^(n - 1) + (n - 1)∫sin^(n - 2) dx - (n - 1)Ln
collecting like terms
n.Ln = -cosx.sin^(n - 1)x + (n - 1)∫sin^(n - 2)x dx
∫sin^n(x)dx = -(1/n)cosx.sin^(n - 1)x + {(n - 1)/n}∫sin^(n - 2)x dx