# Integral of √(a^2 - x^2)

1. Oct 5, 2011

### Masschaos

1. The problem statement, all variables and given/known data
∫√(a2 - x2) dx

2. Relevant equations

3. The attempt at a solution
What I've understood so far is its a substitution integral.
Through the instruction of my lecture notes, I have

∫√(a2 - x2) dx Let u = sin-1(x/a)
x = a*sin(u)
dx = a*cos(u)
This means that √(a2 - x2) = √(a2 - a2sin2(x))

Then as the trig identity of √(cos2(u) + sin2(u)) = 1.
You can rearrange it to give cos(u) = √(1-sin2(u))
So √(a2 - a2sin2(u)) = √(a2 * cos2(u)) = a*cos(u)

Now there is my problem, A lot of other sources say that rather than ∫a*cos(u) du
it is ∫a2 * cos2(u) du
and I don't understand why.

2. Oct 5, 2011

### vela

Staff Emeritus
OK, so you rewrote the radical in terms of u, so at this point you have
$$\int \sqrt{a^2-x^2}\,dx = \int a\cos u\,dx$$Note that the integral still has dx in it, not du.

3. Oct 5, 2011

### Masschaos

So it is ∫a*cos(u) dx and not ∫a2 * cos2 (u) dx?
I'm just struggling as I don't understand why some people are using ∫a2 * cos2 (u) dx.
I've checked on wolfram Alpha also, and there is also the ∫a2 * cos2 (u) dx and I can't reconcile this from my above working.

4. Oct 5, 2011

### daveb

You still have the dx in there. From your first post, what does dx =? Put it in terms of du.

5. Oct 5, 2011

### HallsofIvy

Yes, but you don't want it to be that- you want everything in terms of u, not x.

NO, it isn't! Look again! You cannot change everything except the differential to u but leave the differential as "dx".
With the substitution x= asin(u), dx= d(asin(u))/du du= acos(u)du.

Now, you have both $\sqrt{a^2- x^2}= \sqrt{a^2- a^2sin^2(u)}= \sqrt{a^2cos^2(u)}= a cos(u)$ (as long as we are careful about signs) and $dx= a cos(u)du$. Putting those together, $\int\sqrt{a^2- x^2} dx= \int (a cos(u))(a cos(u)du)= a^2\int cos^2(u)du$.

6. Oct 6, 2011

### Masschaos

Oh of course!
I can't believe I didn't see that.
Thank you very much!