Exploring Substitution Integration for √(a^2 - x^2)

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In summary, the substitution x=asin(u) leads to √(a2-x2) being equal to √(a2-a2sin2(x)) and the attempt at a solution states that cos(u) = √(1-sin2(u)) which results in √(a2-a2sin2(u)) being equal to a*cos(u).
  • #1
Masschaos
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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.
 
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  • #2
Masschaos said:

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
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
[tex]\int \sqrt{a^2-x^2}\,dx = \int a\cos u\,dx[/tex]Note 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.
 
  • #3
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
Masschaos said:
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.
 
  • #5
Masschaos said:
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 [itex]\sqrt{a^2- x^2}= \sqrt{a^2- a^2sin^2(u)}= \sqrt{a^2cos^2(u)}= a cos(u)[/itex] (as long as we are careful about signs) and [itex]dx= a cos(u)du[/itex]. Putting those together, [itex]\int\sqrt{a^2- x^2} dx= \int (a cos(u))(a cos(u)du)= a^2\int cos^2(u)du[/itex].
 
  • #6
Oh of course!
I can't believe I didn't see that.
Thank you very much!
 

What is the integral of √(a^2 - x^2)?

The integral of √(a^2 - x^2) is equal to (1/2)(x√(a^2 - x^2) + a^2sin^-1(x/a)) + C, where C is the constant of integration.

What is the domain and range of the function √(a^2 - x^2)?

The domain of the function √(a^2 - x^2) is [-a, a], as the value under the square root cannot be negative. The range of the function is [0, a], as the output values cannot be negative.

How can the integral of √(a^2 - x^2) be used in real life applications?

The integral of √(a^2 - x^2) can be used in various real life applications, such as calculating the area of a circle with radius a or finding the arc length of a circle with radius a. It is also used in physics and engineering to calculate the work done by a force on an object moving in a circular path.

Can the integral of √(a^2 - x^2) be simplified using trigonometric identities?

Yes, the integral of √(a^2 - x^2) can be simplified using the trigonometric identity sin^2(x) + cos^2(x) = 1. By substituting sin(x) = x/a and cos(x) = √(1 - (x/a)^2), the integral can be simplified to (1/2)(x^2sin^-1(x/a) + a^2sin^-1(x/a)) + C.

What is the derivative of the integral of √(a^2 - x^2)?

The derivative of the integral of √(a^2 - x^2) is equal to √(a^2 - x^2), as the integral is the reverse process of differentiation. This can be verified by taking the derivative of the given integral and simplifying it using the fundamental theorem of calculus.

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