How do you find the indefinite integral of ∫x√9−x2?

2 Answers
Roy E.
Dec 3, 2016

xsin−1(x3)−sin−1(x3)+c.
By parts. Set u(x)=x, dvdx=1√9−x2

Explanation:

Then dudx=1 and v=sin−1(x3).
The integration by parts formula is:
∫u(x)v(x)dx=u(x)v(x)−∫v(x)dudxdx

(or ∫udv=uv−∫vdu if your prefer)

so that v(x)=sin−1(x3)
giving the integral shown in the answer.

Notes:

  1. I assume that either you can quote the standard integral ∫1√a2−x2dx=sin−1(xa) (or cos−1(xa), it makes little difference). If not, you have to remember that you can differentiate all the inverse trig functions e.g. y=sin−1x by getting the x on to the left first: x=siny, then differentiate with respect to y not x, getting dxdy=cosy=√1−x2, then reciprocate both sides dydx=1√1−x2. 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 x2 which is bad news. Similarly if the x had been and x2 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.

mason m
Dec 3, 2016

∫x√9−x2dx=−√9−x2+C

Explanation:

I=∫x√9−x2dx

Let u=9−x2. Differentiating this shows that du=−2xdx. Luckily our numerator is this only off by a factor of −2.

I=−12∫−2x√9−x2dx=−12∫1√udu=−12∫u−12du

Using ∫undu=un+1n+1+C, this becomes:

I=−12(u1212)=−√u=−√9−x2+C

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