Question

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

Answer 1

See the explanation.

Explanation:

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

`dv=dx => v=x`

`I_n=int (arcsinx)^ndx = x(arcsinx)^n - int x(n(arcsinx)^(n-1))/sqrt(1-x^2)dx`

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

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

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

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

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

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

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

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

So, we have recurrent formula, we only have to find `I_1` and `I_2` and that's trivial.