Question

How do you find the integral of `(sin^(-1)x)^2`?

Answer 1

`int(sin^-1x)^2dx=x(sin^-1x)^2+2sqrt(1-x^2)sin^-1x-2x+C`

Explanation:

`I=int(sin^-1x)^2dx`

We will use integration by parts. This takes the form `intudv=uv-intvdu`. For the given integral `I`, let:

`{(u=(sin^-1x)^2),(dv=dx):}`

Differentiating `u` and integrating `dv` show:

`{(du=(2sin^-1x)/sqrt(1-x^2)dx),(v=x):}`

Note that differentiating `u` requires the chain rule.

Then:

`I=uv-intvdu=x(sin^-1x)^2-int(2x(sin^-1x))/sqrt(1-x^2)dx`

Perform integration by parts again. Note that the portion `int(2x)/sqrt(1-x^2)dx` is easily integrable with the substitution `t=1-x^2=>dt=-2xdx`, so:

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

Then for integration by parts let:

`{(u=sin^-1x" "=>" "du=1/sqrt(1-x^2)dx),(dv=(2x)/sqrt(1-x^2)dx" "=>" "v=-2sqrt(1-x^2)):}`

So:

`I=x(sin^-1x)^2-(uv-intvdu)`

`I=x(sin^-1x)^2-uv+intvdu`

`I=x(sin^-1x)^2-sin^-1x(-2sqrt(1-x^2))+int(-2sqrt(1-x^2))/sqrt(1-x^2)dx`

`I=x(sin^-1x)^2+2sqrt(1-x^2)sin^-1x-2intdx`

`I=x(sin^-1x)^2+2sqrt(1-x^2)sin^-1x-2x+C`