Integrate sqrt(9-x^2) by hand

1. Jan 17, 2013

physicsdreams

1. The problem statement, all variables and given/known data

Evaluate
$$\int_ \ \sqrt[2]{9-x^{2}} \, \mathrm{d} x.$$

2. Relevant equations

U substitution
Integration by parts

3. The attempt at a solution

My teacher said it wasn't possible "by hand", but Wolframalpha provided a step-by-step solution. I'm clueless as to how wolfram came up with the substitutions/steps and how a mere mortal as myself could possibly be expected to do the same. If you could walk me through the logic behind the steps that wolfram provides, I would greatly appreciate it. I'm really just curious about the process as I just can't seem to wrap my head around it.

Wolfram's steps: http://www.wolframalpha.com/widgets/view.jsp?id=dc816cd78d306d7bda61f6facf5f17f7

Thanks!

Last edited: Jan 17, 2013
2. Jan 17, 2013

Dick

What do you mean 'by hand'. You can certainly integrate it with a trig substitution. Try x=3*sin(t). What's dx? The Wolfram reference you gave is about integrating sec^3(x). I don't see what that has to do with this problem.

Last edited by a moderator: Jan 17, 2013
3. Jan 17, 2013

physicsdreams

Hey Dick,

Thanks for your reply. I suppose we haven't learned trig substitution yet, so I guess that's my problem.
As for the dx, doesn't that mean 'in respect to x'?

Thanks!

4. Jan 17, 2013

Dick

Not knowing trig substitution probably is not a help. But you know ordinary u substitution, yes? It's the same thing except you use trig functions. If you put x=t^2 then dx=2tdt. If you put x=3*sin(t) then dx=3*cos(t)dt. Same general idea.

5. Jan 17, 2013

physicsdreams

Alright then, I guess that's something I'll have to study later on.
Thanks!

6. Jan 17, 2013

Dick

Well, ok. You could probably do it now if you took a crack at it. It's not that mysterious. But if you want to wait till later, that's ok too.

7. Jan 17, 2013

Staff: Mentor

If it's a definite integral, with certain limits, then you can evaluate the integral be exploiting the geometry involved.

For example, if it happens to be
$$\int_0^3 \sqrt{9 - x^2}dx$$
then the integral represents one quarter of the area of a circle of radius 3, centered at (0, 0).

8. Jan 18, 2013

lurflurf

Integrate by parts

$$\int \sqrt{a^2-x^2} \mathrm{dx}= x\sqrt{a^2-x^2}-\int \frac{-x^2}{\sqrt{a^2-x^2}} \mathrm{dx}$$
$$=x \sqrt{a^2-x^2}-\int \frac{a^2-x^2}{\sqrt{a^2-x^2}} \mathrm{dx}+a^2 \int \frac{1}{\sqrt{a^2-x^2}} \mathrm{dx}=x \sqrt{a^2-x^2}-\int \sqrt{a^2-x^2} \mathrm{dx}+a^2\arcsin(x/a)$$
From which we conclude
$$\int \sqrt{a^2-x^2} \mathrm{dx}=\frac{1}{2}\left(x\sqrt{a^2-x^2}+a^2\arcsin(x/a)\right)$$
An unrestricted form is
$$\int \sqrt{a^2-x^2} \mathrm{dx}=\frac{1}{2}\left(x \sqrt{a^2-x^2}+\arctan \left(\frac{x}{\sqrt{a^2-x^2}}\right)\right)$$

Last edited by a moderator: Jan 18, 2013