Question

Show that `I_n = -1/n \ cosx \ sin^(n-1)x+(n-1)/n \ I_(n-2)` where `I_n=int \ sin^nx \ dx`, and `n ge 2`?

Answer 1

We want to show that:

`I_n = -1/n \ cosx \ sin^(n-1)x+(n-1)/n \ I_(n-2)`

Where `I_n=int \ sin^nx \ dx`, and `n ge 2`.

We can write the integral as:

`I_n=int \ (sinx)(sin^(n-1)x) \ dx` provided `n ge 2`

We can use integration by parts:

Let `{ (u,=sin^(n-1)x, => (du)/dx,=(n-1)sin^(n-2)xcosx), ((dv)/dx,=sinx, => v,=-cosx ) :}`

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

gives us:

`int \ (sin^(n-1)x)(sinx) \ dx = (sin^(n-1)x)(-cosx) - int \ (-cosx)((n-1)sin^(n-2)xcosx) \ dx`

`:. I_n = -cosx \ sin^(n-1)x + (n-1)int \ cos^2x \ sin^(n-2)x \ dx`

Using the identity, `sin^2A+cos^2A -=1` we have:

`I_n = -cosx \ sin^(n-1)x + (n-1)int \ (1-sin^2x) \ sin^(n-2)x \ dx`
`\ \ \ = -cosx \ sin^(n-1)x + (n-1)int \ sin^(n-2)x - sin^(n)x \ dx`
`\ \ \ = -cosx \ sin^(n-1)x + (n-1)(I_(n-2) - I_n)`
`\ \ \ = -cosx \ sin^(n-1)x + (n-1)I_(n-2) - (n-1)I_n`

`I_n + (n-1) I_n = -cossx \ sin^(n-1)x + (n-1)I_(n-2)`

`:. nI_n = -cosx \ sin^(n-1)x + (n-1)I_(n-2)`

`:. I_n = -1/ncossx \ sin^(n-1)x + (n-1)/nI_(n-2)` QED