Question

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` ?

Answer 1

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`