Question

How do you integrate `int (sin^-1x)^2` using integration by parts?

Answer 1

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

Explanation:

We have:

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

Integration by parts takes the form `intudv=uv-intvdu`. For `int(sin^-1x)^2dx`, we have to choose values of `u` and `dv`.

Whatever value of `dv` has to be integrated, so it would be foolish to choose `sin^-1x` or `(sin^-1x)^2` as `dv` because their integrals are not clear. So, let `u=(sin^-1x)^2` and `dv=dx`, all that remains.

So, we have:

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

Take the derivative of `u` and integrate `dv`:

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

So we see that:

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

Or:

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

We now have another integral to use integration by parts on. Again, don't choose `sin^-1x` as `dv`, so let `dv` be everything else.

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

Now differentiate and integrate, respectively. Note that `intx/sqrt(1-x^2)dx` can be performed with the substitution `t=1-x^2`.

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

Then:

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

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

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