How do you use integration by parts to establish the reduction formula #intsin^n(x) dx = -(1/n)sin^(n-1)(x)cos(x)+(n-1)/nintsin^(n-2)(x)dx# ?

Question

3733 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-09-19T20:38:19

Let #I=int sin^nx dx#.

By pulling out one of #sin x#'s,

#I=int sinx cdot sin^{n-1}x dx#

Let #u=sin^{n-1}x# and #dv=sinx dx#
#Rightarrow du=(n-1)sin^{n-2}xcosx# and #v=-cosx#

by Integration by Parts,

#=-sin^{n-1}xcosx+(n-1)int sin^{n-2}xcos^2xdx#

by the trig identity #cos^2x=1-sin^2x#,

#=-sin^{n-1}xcosx+(n-1)int sin^{n-2}x(1-sin^2x)dx#

#=-sin^{n-1}xcosx+(n-1)int sin^{n-2}xdx-(n-1)I#

By adding #(n-1)I#

#Rightarrow nI=-sin^{n-1}xcosx+(n-1)int sin^{n-2}xdx#

By dividing by #n#,

#I=-1/nsin^{n-1}xcosx+(n-1)/nint sin^{n-2}xdx#