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Integrate sqrt ( 4 -x^2)dx

  1. Apr 1, 2005 #1
    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.
     
  2. jcsd
  3. Apr 1, 2005 #2
    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.
     
  4. Apr 1, 2005 #3
    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.
     
  5. Apr 1, 2005 #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!
     
  6. Apr 1, 2005 #5

    dextercioby

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

    Daniel.
     
  7. Apr 1, 2005 #6
    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]
     
    Last edited: Apr 1, 2005
  8. Apr 1, 2005 #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.
     
  9. Apr 1, 2005 #8
    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.
     
  10. Apr 1, 2005 #9

    dextercioby

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    And the last integral can be calculated again by parts...

    Daniel.
     
  11. Apr 1, 2005 #10
    Right. I was hoping the original poster would comment again before I did too many steps.
     
  12. Apr 1, 2005 #11
    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)....
     
  13. Apr 1, 2005 #12
    You know that [tex]\sin{\theta}=\sqrt{1-\cos^2{\theta}}[/tex], right? Does that help?
     
  14. Apr 2, 2005 #13
    then. theta for the second part would be

    theta = arcsin sqrt ( 1- x^2/4).....
     
  15. Apr 2, 2005 #14
    [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]
     
  16. Apr 2, 2005 #15
    Is Halls of Ivy method wrong then
     
  17. Apr 2, 2005 #16

    dextercioby

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    Of course not.

    Daniel.
     
  18. Apr 3, 2005 #17
    [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?
     
  19. Apr 4, 2005 #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.
     
  20. Apr 5, 2005 #19
    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.
     
  21. Apr 5, 2005 #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.
     
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