Question

How do you find `\int \sin ^ { - 1} x d x`?

Answer 1

`intsin^-1(x)dx=xsin^-1(x)+sqrt(1-x^2)+C`

Explanation:

We have:

`I=intsin^-1(x)dx`

In the absence of being able to do anything else, we should try to use integration by parts.

Integration by parts says to let the given integral equal to `intudv`, which is then equal to `uv-intvdu`.

`intudv=uv-intvdu`

So, we want to choose a `u` value that will get simpler when we differentiate and a `dv` value that we can easily integrate.

Since all we have are `sin^-1(x)` and `dx` in the integral, let `u=sin^-1(x)`, since if this were part of `dv` we would be stuck. Thus let `dv=dx`, the only thing that remains.

If we have `color(red)(u=sin^-1(x)` then `color(red)(du=1/sqrt(1-x^2)dx`.

From `color(blue)(dv=dx` we see that `color(blue)(v=x`.

Then:

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

The second integral can be found through examination or through the substitution `t=1-x^2`.

`I=xsin^-1(x)+1/2int(1-x^2)^(-1/2)(-2xdx)`

`I=xsin^-1(x)+1/2intt^(-1/2)dt`

`I=xsin^-1(x)+1/2(2t^(1/2))`

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