Question

How do you find the indefinite integral of `int x/(sqrt(9-x^2))`?

Answer 1

`x sin^{-1}(x/3)-sin^{-1}(x/3)+c`.
By parts. Set `u(x)=x`, `{dv}/{dx}=1/sqrt(9-x^2)`

Explanation:

Then `{du}/{dx}=1` and `v=sin^{-1}(x/3)`.
The integration by parts formula is:
`int u(x)v(x) dx = u(x)v(x)-int v(x){du}/{dx}dx`

(or `int udv = uv-int vdu` if your prefer)

so that `v(x)=sin^{-1}(x/3)`
giving the integral shown in the answer.

Notes:

1. I assume that either you can quote the standard integral `int 1/sqrt(a^2-x^2)dx=sin^{-1} (x/a)` (or `cos^{-1}(x/a)`, it makes little difference). If not, you have to remember that you can differentiate all the inverse trig functions e.g. `y=sin^{-1}x` by getting the `x` on to the left first: `x=sin y`, then differentiate with respect to `y` not `x`, getting `dx/{dy}=cos y = sqrt(1-x^2)`, then reciprocate both sides `dy/{dx}=1/(sqrt(1-x^2)`. Then the integration is reverse of differentiation.

2. You have to guess correctly which part to make `u(x)` and which to make `v(x)`. A simple rule of thumb here is that that the `x` on the top is an irritation and so if you differentiate it will turn into 1 and go away. If you guess wrong the `x` will turn into an `x^2` which is bad news. Similarly if the `x` had been and `x^2` you would have needed two rounds of integration by parts.

3. I gloss over certain issues about "principal value" and the choice of `sin^-1x` or `cos^-1x`.

Answer 2

`intx/sqrt(9-x^2)dx=-sqrt(9-x^2)+C`

Explanation:

`I=intx/sqrt(9-x^2)dx`

Let `u=9-x^2`. Differentiating this shows that `du=-2xdx`. Luckily our numerator is this only off by a factor of `-2`.

`I=-1/2int(-2x)/sqrt(9-x^2)dx=-1/2int1/sqrtudu=-1/2intu^(-1/2)du`

Using `intu^ndu=u^(n+1)/(n+1)+C`, this becomes:

`I=-1/2(u^(1/2)/(1/2))=-sqrtu=-sqrt(9-x^2)+C`