How do you solve the integral ∫√(a² - x²) dx using substitution?

[Masschaos]

Homework Statement

∫√(a² - x²) dx

Homework Equations

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.

 

[vela]

Homework Statement

∫√(a² - x²) dx

Homework Equations

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) du[/color]
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)

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\,dxNote that the integral still has dx in it, not du.

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.

 

[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.

 

[daveb]

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.

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

 

[HallsofIvy]

So it is ∫a*cos(u) dx and not ∫a² * cos² (u) dx?

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

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.

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.

 

[Masschaos]

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