Integrate sqrt ( 4 -x^2)dx

  • Thread starter Naeem
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  • #1
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Hi,

Can anyone help me integrate :

Integral sqrt ( 4 - x^2) dx

Some ideas:

Make a subsitution x2 = cos (theta)

Taking the derivative of both sides:

2x = - sin (theta).d(theta)

or....

Double angle identity:

(Cosx)^2 = 1 + cos2x / 2


Can anybody give some ideas.
 

Answers and Replies

  • #2
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Trigonometrically: x = 2cos(theta), or 2sin(theta), not [tex] x^2 = 2cos(theta) [/tex]

You cant use the double angle identity without making that substitution, so its not really 'or' is it :)

You might be able to do it by parts with u = sqrt(4-x^2) and dv = dx.
 
  • #3
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By parts will work.

[tex]\int udv = uv - \int{vdu}[/tex]

Let u = [itex]\sqrt{4-x^2}[/itex]
And du = [tex]\frac{2x}{2\sqrt{4-x^2}}[/tex]

Let dv = dx
And v = x

[tex]\int \sqrt{4-x^2}dx = x\sqrt{4-x^2} - \int\frac{2x^2}{2\sqrt{4-x^2}}dx[/tex]

Hope that helps.
 
  • #4
HallsofIvy
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I don't see any reason for a double angle. If you factor out that 4 you get
[tex]\sqrt{4- x^2}= 2\sqrt{1- \frac{x^2}{4}[/tex]. And anytime you see something like that you should think "[tex]\sqrt{1- cos^2x}= sin x[/tex]" (or "[tex]\sqrt{1- sin^2x}= cos x[/tex]").

Let [tex]cos(\theta)= \frac{x}{2}[/tex] or [2 cos(\theta)= x[/tex] so that
[tex]-2sin(\theta)dx= dx[/tex]. Then [tex]\integral \sqrt{4-x^2}dx= 2\integral \sqrt{1- (\frac{x}{2})^2}dx= -4\integral\sqrt{1- cos^2\theta}sin \theta d\theta= -4\integral sin^2 \theta d\theta[/tex].

That last integral you can do using [tex]sin^2 \theta= \frac{1}{2}(1- cos \theta)[/tex].
Well, by golly, you do need a double angle!
 
  • #5
dextercioby
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Jameson said:
By parts will work.

[tex]\int udv = uv - \int{vdu}[/tex]

Let u = [itex]\sqrt{4-x^2}[/itex]
And du = [tex]\frac{2x}{2\sqrt{4-x^2}}[/tex] $
Let dv = dx
And v = x

[tex]\int \sqrt{4-x^2}dx = x\sqrt{4-x^2} - \int\frac{2x^2}{2\sqrt{4-x^2}}dx[/tex]

Hope that helps.
U missed a "-" (=minus) when computing the marked differential.Not to mention the "dx".

Daniel.
 
  • #6
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You're right. The "dx's" aren't as critical as the minus was... here's the corrected integral.
------------------
[tex]\int udv = uv - \int{vdu}[/tex]

Let u = [itex]\sqrt{4-x^2}[/itex]
And du = [tex]\frac{-2x}{2\sqrt{4-x^2}}dx[/tex]

Let dv = dx
And v = x

[tex]\int \sqrt{4-x^2}dx = x\sqrt{4-x^2} + \int\frac{x^2}{\sqrt{4-x^2}}dx[/tex]
 
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  • #7
dextercioby
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Apparently u forgot about it,again.It should have been

[tex] (...)+\int \frac{x^{2}}{\sqrt{4-x^{2}}} \ dx [/tex]

,where the 2-s got simplified through.

Daniel.
 
  • #8
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Oh, alright. Sorry I didn't see it before. I didn't catch what you were referring to. I'll edit my previous post... hope I get it right this time.
 
  • #9
dextercioby
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And the last integral can be calculated again by parts...

Daniel.
 
  • #10
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Right. I was hoping the original poster would comment again before I did too many steps.
 
  • #11
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I did this problem using Halls of Ivy idea, i.e substitution, and I got

Integral sqrt( 4 -x^2)dx = -2 Integral ( 1-cos ( theta ) dtheta =
= -2[ Integral dtheta - Integral cos(theta)dtheta ]
= -2 ( theta - sin (theta) + C
= -2 ( cos-1 (x/2) - sin ........ what do I put theta as here)....
 
  • #12
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Naeem said:
= -2 ( cos-1 (x/2) - sin ........ what do I put theta as here)....
You know that [tex]\sin{\theta}=\sqrt{1-\cos^2{\theta}}[/tex], right? Does that help?
 
  • #13
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then. theta for the second part would be

theta = arcsin sqrt ( 1- x^2/4).....
 
  • #14
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[tex]\int \sqrt{4-x^2}dx [/tex]

[tex] u = \sqrt{4-x^2}, du = \frac{-x}{\sqrt{4-x^2}}dx, v = x, dv = dx [/tex]

[tex]\int \sqrt{4-x^2}dx = x\sqrt{4-x^2} + \int\frac{x^2}{\sqrt{4-x^2}}dx[/tex]

x = 2sin(t)
dx = 2cos(t) dt

The integral on the RHS becomes:

[tex] \int \frac{8cos(t)sin^2(t)}{\sqrt{4-4sin^2(t)}}dt = \int \frac{8cos(t)sin^2(t)}{2\sqrt{cos^2(t)}}dt = \int \frac{8cos(t)sin^2(t)}{2cos(t)}}dt = 4\int{sin^2(t)}{dt}[/tex]

From there use the trig identity[tex] sin^2(t) = \frac{1-2cos(t)}{2} [/tex]

[tex] 2\int{1-cos(2t)}{dt} = 2t - 2\int{cos2t}{dt} = 2t - sin(2t)[/tex]

From x = 2sin(t), t = arcsin(x/2).

[tex]2(arcsin(x/2)-sin(2)) [/tex]

Since sin(2) is constant it can be omitted and

[tex] \int \sqrt{4-x^2}dx = x\sqrt{4-x^2} + 2arcsin(x/2) [/tex]
 
  • #15
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Is Halls of Ivy method wrong then
 
  • #17
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[tex] \int \sqrt{4-x^2}dx = x\sqrt{4-x^2} + 2arcsin(x/2) [/tex]

When I evaluate the integral on the LHS on maple it tells me that the first term on the RHS should be divided by 2.

Can anyone figure this out?
 
  • #18
dextercioby
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Maple is correct (up to a constant he never considers :wink:).

[tex] \int \sqrt{4-x^{2}} \ dx =2 \arcsin\left(\frac{x}{2}\right)+2\frac{x}{2}\sqrt{1-\left(\frac{x}{2}\right)^{2}} + C =2 \arcsin\left(\frac{x}{2}\right)+\frac{1}{2}x\sqrt{4-x^{2}} + C [/tex]

Daniel.
 
  • #19
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So you square the 1/2 and put it inside making it 1-(x/2)^2. But why does it still have 4 - x^2?

You lost me.
 
  • #20
dextercioby
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Well,when reversing the substitution u end up with

[tex] 2\frac{x}{2}\sqrt{1-\left(\frac{x}{2}\right)^{2}} [/tex]

which used to stand for [itex] \sin 2u =2 \sin u \ \cos u [/tex]

U know algebra,so how about work that square & square roots...?

Daniel.
 
  • #21
HallsofIvy
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The fact that there are different ways of doing a problem does not mean that any of them is wrong!

The original problem was Integral sqrt ( 4 - x^2) dx [tex]\integral \sqrt{4-x^2}dx[/tex]

Let [tex]2 sin(\theta)= x[/tex] so that 2 cos(\theta)= dx. [tex]\sqrt{4- x^2}= \sqrt{4- 4sin^2(\theta)}= 2 cos(\theta)[/tex]. The integral becomes [tex]4 \int cos^2(\theta)d\theta= 2 \int (1+ cos(2\theta))d\theta = 2\theta+ sin(2\theta)[/tex].
Now, go back to x.
 
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  • #22
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No, it's fine.

Here:

[tex]\int \sqrt{4-x^2} \ dx = \int -2\sin \theta \sqrt{4 - 4\cos^2 \theta} \ dx = -4 \int \sin^2 \theta \ dx = -2\int 1 - \cos 2\theta \ d\theta = -2\theta + \sin 2\theta + C = -2\arccos \left( \frac{x}{2} \right) + 2\sin \theta \cos \theta+C [/tex]

[tex]= \frac{x}{2}\sqrt{4-x^2} - 2\arccos \left( \frac{x}{2} \right)+C [/tex]

this is an equivalent answer (it differs only from those above by a constant). Note that here [itex]2\cos \theta = x[/itex].
 
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  • #23
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oops, didn't notice the second page!
 
  • #24
dextercioby
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Here's another way to do it,if u have the patience (:tongue2:).A mathematician should.

Make the obvious substitution

[tex] \frac{x}{2}=\tanh u [/tex]

and go from there...

Daniel.
 
  • #25
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[tex]\int \sqrt{4-x^2}dx [/tex]

[tex] u = \sqrt{4-x^2}, du = \frac{-x}{\sqrt{4-x^2}}dx, v = x, dv = dx [/tex]

[tex]\int \sqrt{4-x^2}dx = x\sqrt{4-x^2} + \int\frac{x^2}{\sqrt{4-x^2}}dx[/tex]

x = 2sin(t)
dx = 2cos(t) dt

The integral on the RHS becomes:

[tex] \int \frac{8cos(t)sin^2(t)}{\sqrt{4-4sin^2(t)}}dt = \int \frac{8cos(t)sin^2(t)}{2\sqrt{cos^2(t)}}dt = \int \frac{8cos(t)sin^2(t)}{2cos(t)}}dt = 4\int{sin^2(t)}{dt}[/tex]

From there use the trig identity[tex] sin^2(t) = \frac{1-2cos(t)}{2} [/tex]

[tex] 2\int{1-cos(2t)}{dt} = 2t - 2\int{cos2t}{dt} = 2t - sin(2t)[/tex]

From x = 2sin(t), t = arcsin(x/2).

[tex]2(arcsin(x/2)-sin(2)) [/tex]

Since sin(2) is constant it can be omitted and

[tex] \int \sqrt{4-x^2}dx = x\sqrt{4-x^2} + 2arcsin(x/2) [/tex]

at the very end, how come sin2t becomes a sin2?
If you substitute t=arcsin(x/2) into sin2t, don't we get sin(2arcsin(x/2))?
how could that come out to be sin2?

Can someone plz help me? THX!!!
 

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