Integrating arcsin: Solving \int \sqrt{9-x^{2}}dx with step-by-step explanation

[Hannisch]

Homework Statement

The problem, from the very beginning, was:

$$\int \sqrt{9-x^{2}}dx$$

But this I have reduced to:

$$\int \sqrt{9-x^{2}}dx = \frac{x}{2} \sqrt{9-x^{2}} + \frac{9}{2} \int \frac{1}{\sqrt{9-x^{2}}}dx$$

My problem is that last integral - I get a factor of (1/3) times the correct answer and I don't know what to do - I simply can't see it.

Homework Equations

$$\int \frac{1}{\sqrt{1-x^{2}}}dx = arcsinx$$

The Attempt at a Solution

I look at it and want to "transform" my expression into something like the arcsin expression above. So I say:

$$\int \frac{1}{\sqrt{9-x^{2}}}dx = \int \frac{1}{3\sqrt{1-\frac{x^{2}}{9}}}dx$$

and from there get:

$$\int \frac{1}{\sqrt{9-x^{2}}}dx = \frac{1}{3}arcsin\frac{x}{3}$$

but you're not supposed to get that factor of 1/3 - you're not supposed to remove it? Can anyone explain to me what I'm missing? I've searched my textbook so many times now I'm about to throw it into the wall or something..

 

[Donaldos]

$$\int \frac{{\rm d}u}{\sqrt{1-u^2}}=\arcsin u$$

but here we have

$$\int \frac{{\rm d}x}{\sqrt{1-\left(\frac x 3 \right)^2}}$$

$$u= \frac x 3 \Rightarrow {\rm d}x= ?$$

 

[scottie_000]

You need to use that relevant equation to deal with $$\frac{x}{3}$$ and not just $$x$$. Be careful with that substitution!

Also, a more natural way to deal with the original integral would be to consider the substitution $$x=3 \sin \theta$$ and go from there using some trig identities

 

[Hannisch]

Ohh, I think I get it now, thank you! I'd say it needs dx/3, so my 1/3 "disappears" (for lack of better wording).

Anyway, as for that last comment - you're probably right, but I don't think I've seen that type of thing being used before.

 

[scottie_000]

Anyway, as for that last comment - you're probably right, but I don't think I've seen that type of thing being used before.

No need to worry really. You will come across it sooner or later and it's good to know

 

[Bohrok]

You could also have used the trig substitution x = 3sinθ which should make the integral pretty easy to work with.