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Integration of (R^2 - x^2)^.5

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data
    How can you integrate the following?

    [tex] \int \sqrt {R^2 - x^2} dx [/tex]


    3. The attempt at a solution

    Integration by parts does not seem to be a solution here.

    Integration by substitution works to some extend:

    Let [itex] x = R sin(u) [/itex].

    We get
    [tex] \int \sqrt {R^2 - x^2} dx = \int Rcos(u) (R^2 + 2R^2 sin^2(u)) Rsin(u) du[/tex]
    [tex] = \int R^4 cos^2(u) sin(u) (1 + 2sin^2(u) ) du [/tex]

    I cannot see how to integrate this easily.
    Perhaps, there is a better way to do this.
     
  2. jcsd
  3. Oct 28, 2009 #2
    Seems like you've got the right idea. Solve this using trigonometric substitution:

    1) Draw a right triangle. Label one of the vertices x, label the hypotenuse R.

    Label the angle [tex]\vartheta[/tex] so [tex]sin(\vartheta)=x/R[/tex].

    2) Substitution

    [tex]x = Rsin(\vartheta), dx = Rcos(\vartheta)d\vartheta[/tex]

    Note from your diagram:
    [tex]\frac{\sqrt{R^{2} - x^{2}}}{R} = cos(\vartheta)[/tex]

    So [tex]\sqrt{R^{2} - x^{2}} = Rcos(\vartheta)[/tex]

    3) With that figured out your integral should be much simpler!
     
    Last edited: Oct 28, 2009
  4. Oct 28, 2009 #3
    You are right.
    I get

    [tex] \frac{1}{2} R^2 \int sin(2\vartheta) = R^2 cos(2\vartheta) [/tex]
     
  5. Oct 28, 2009 #4
    You evaluated

    [tex]\int x \, dx [/tex]

    Whereas your original integral was

    [tex]\int \sqrt {R^2 - x^2} dx [/tex]
     
  6. Oct 28, 2009 #5
    You need to evaluate

    [tex]\int R^{2}cos^{2}(\vartheta) \, d\vartheta [/tex]

    Make sure you after you solve the integral you substitute back for the solution in terms of x.
     
  7. Oct 28, 2009 #6
    I get this

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

    It should be correct, since SageMath gives the same result.
     
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