indefinite integral of x^2.arcsin(x)

Question

find the indefinite integral of the product of x^2 and the inverse sin of x, that is [x^2.sin^(-1)x], using integration by parts

Answer 1

indefinite integral of x^2.arcsin(x), using integration by parts.

we use the fact that int asin(x) = x.asin(x) + (1-x^2)^(1/2)

int x^2.arcsin(x) dx = x^2*{x.asin(x) + (1-x^2)^(1/2)} - {int 2x.x.asin(x) + 2x.(1-x^2)^(1/2) dx}

int x^2.arcsin(x) dx = x^2*{x.asin(x) + (1-x^2)^(1/2)} - 2{int x^2.asin(x) dx} - {int 2x.(1-x^2)^(1/2) dx}

3{int x^2.arcsin(x) dx} = x^2*{x.asin(x) + (1-x^2)^(1/2)} - {int 2x.(1-x^2)^(1/2) dx}

3{int x^2.arcsin(x) dx} = x^2*{x.asin(x) + (1-x^2)^(1/2)} - {-(2/3)(1-x^2)^(3/2)}

3{int x^2.arcsin(x) dx} = x^2.x.asin(x) + x^2.(1-x^2)^(1/2) + (2/3)(1-x^2)^(3/2)

3{int x^2.arcsin(x) dx} = x^3.asin(x) + (1-x^2)^(1/2){x^2 + (2/3)(1-x^2)}

int x^2.arcsin(x) dx = (1/3).x^3.asin(x) + (1/9).{3x^2 + 2(1-x^2)}.(1-x^2)^(1/2)

int x^2.arcsin(x) dx = (1/3).x^3.asin(x) + (1/9).{x^2 + 2x^2)^(1/2)