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

Question

9666 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-03T15:56:07

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

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:

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

Answer 2 · 2016-12-03T21:44:31

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

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

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:

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

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