Question

How do you find the integral of `int [arccos(x)]^n` if is an integer?

Answer 1

`I_n = x(arccosx)^n - n(arccosx)^(n-1)sqrt(1-x^2) - n(n-1)I_(n-2)`

Explanation:

`u=(arccosx)^n => du=n(arccosx)^(n-1)*(-1/sqrt(1-x^2))dx`

`dv=dx => v=x`

`I_n = x(arccosx)^n - int x*n(arccosx)^(n-1)*(-1/sqrt(1-x^2))dx`

`I_n = x(arccosx)^n + nint (arccosx)^(n-1)*x/sqrt(1-x^2)dx`

`u=(arccosx)^(n-1) => du=(n-1)(arccosx)^(n-2)*(-1/sqrt(1-x^2))dx`

`dv = x/sqrt(1-x^2)dx`

`v = int x/sqrt(1-x^2)dx`

`1-x^2=t => -2xdx=dt => xdx = -dt/2`

`v = -1/2int dt/sqrtt = -1/2 t^(1/2)/(1/2) = -sqrtt = -sqrt(1-x^2)`

`I_n = x(arccosx)^n + n[-(arccosx)^(n-1)sqrt(1-x^2) - int ( -sqrt(1-x^2))(n-1)(arccosx)^(n-2)*(-1/sqrt(1-x^2))dx]`

`I_n = x(arccosx)^n + n[-(arccosx)^(n-1)sqrt(1-x^2) - (n-1) int (arccosx)^(n-2)dx]`

`I_n = x(arccosx)^n - n(arccosx)^(n-1)sqrt(1-x^2) - n(n-1)I_(n-2)`

We have recurrent formula, we only have to find `I_1` and `I_2` and that's trivial using Integration By Parts.