`int (arcsin(x))^2 dx` Evaluate the integral

You need to use integration by parts, such that:

`int udv = uv - int v du`

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

`dv = 1 => v = x`

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

You need to solve the integral `int arcsin x*(x/(sqrt(1-x^2)))dx` using substitution `arcsin x= t` , such that:

`arcsin x = t => (dx)/(sqrt(1-x^2))= dt`

`int arcsin x*(x/(sqrt(1-x^2)))dx = int t*sin t dt`

You need to use...

You need to use integration by parts, such that:

`int udv = uv - int v du`

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

`dv = 1 => v = x`

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

You need to solve the integral `int arcsin x*(x/(sqrt(1-x^2)))dx` using substitution `arcsin x= t` , such that:

`arcsin x = t => (dx)/(sqrt(1-x^2))= dt`

`int arcsin x*(x/(sqrt(1-x^2)))dx = int t*sin t dt`

You need to use integration by parts to solve `int t*sin t dt` , such that:

`u = t => du = dt`

`dv = sin t => v = -cos t`

`int t*sin t dt = -t*cos t + int cos t dt`

`int t*sin t dt = -t*cos t + sint  + c`

Replacing back the variable yields:

`int arcsin x*(x/(sqrt(1-x^2)))dx = -arcsin x*cos(arcsin x) + x + c`

`int (arcsin^2(x)) dx = x*(arcsin^2(x)) - 2( -arcsin x*cos(arcsin x) + x) + C`

`int (arcsin^2(x)) dx = x*(arcsin^2(x)) + 2arcsin x*cos(arcsin x) - 2x + c`

Hence, evaluating the indefinite integral, using parts and substitution, yields `int (arcsin^2(x)) dx = x*(arcsin^2(x)) + 2arcsin x*cos(arcsin x) - 2x + c.`