Question

If `I_n=int_(pi/4)^(pi/2) cot^n x dx`, then prove that `(n-1)(I_n+I_(n-2))=1`?

Answer 1

We have:

`I_n = int_(pi/4)^(pi/2) cot^n x \ dx`

Define `J_n` by:

`J_n = I_n + I_(n-2)`

Then we have:

`J_n = int_(pi/4)^(pi/2) cot^n x \ dx + int_(pi/4)^(pi/2) cot^(n-2) x \ dx`
`\ \ \ = int_(pi/4)^(pi/2) cot^n x + cot^(n-2) x \ dx`
`\ \ \ = int_(pi/4)^(pi/2) cot^(n-2) xcot^2x + cot^(n-2) x \ dx`
`\ \ \ = int_(pi/4)^(pi/2) cot^(n-2) x(cot^2x + 1) \ dx`
`\ \ \ = int_(pi/4)^(pi/2) cot^(n-2) xcsc^2x \ dx`

We can integrate this using Integration By Parts:

Let `{ (u,=cot^(n-2) x , => (du)/dx=(n-2)cot^(n-3)x(-csc^2x)), ((dv)/dx,=csc^2x, => v=-cotx ) :}`

Then plugging into the IBP formula:

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

gives us

`int_(pi/4)^(pi/2) (cot^(n-2) x)(csc^2x) \ dx = [(cot^(n-2) x)(-cotx)]_(pi//4)^(pi//2) - int_a^(pi/2) (-cotx)((n-2)cot^(n-3)x(-csc^2x)) \ dx`

`:. J_n = [cot^(n-2) x(-cotx)]_(pi/4)^(pi/2) - (n-2)int_(pi/4)^(pi/2) cot^(n-2)xcsc^2x dx`

`:. J_n = [-cot^(n-1) x]_(pi//4)^(pi//2) - (n-2)J_n`

`:. J_n +(n-2)J_n= -[cot^(n-1) x]_(pi//4)^(pi//2)`

`:. (n-1)J_n= -(cot^(n-1) (pi/2) - cot^(n-1) (pi/4))`

`:. (n-1)J_n= -(cot^(n-1) (pi/2) - cot^(n-1) (pi/4))`

`:. (n-1)J_n= -(0 - 1)`

`:. (n-1)J_n= 1`

And then by definition

`(n-1) (I_n + I_(n-2))= 1` QED