integration of sin^n x.dx

Question

answer it in terms of n so that by putting any value of n we can get that answer

Answer 1

∫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