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Integrating the function √(12 -x^2)

  1. Oct 21, 2011 #1
    1. The problem statement, all variables and given/known data

    Integrate the function,

    f(x) = √(12 -x2)


    2. Relevant equations

    n/a

    3. The attempt at a solution

    I tried splitting the function up as follows:

    f(x) = √(12+x)*√(12-x)

    then I tried substituting in,

    w=12-x and dw=-dx, to get,

    ∫-2w3/2(2√12-w)1/2dw

    and finally I attempted to use integration by parts on the above integral; however this seems to be a dead end. I know the answer is:

    1/2*√(-x^2 + 12)*x + 6*arcsin(1/6*sqrt(3)*x)

    I just can't seem to work it out for myself.

    Any help would be greatly appreciated.

    Regards,

    Chris
     
  2. jcsd
  3. Oct 21, 2011 #2

    SammyS

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    For [itex]\displaystyle \int\sqrt{12-x^2}\,dx[/itex] use the substitution x = (√12)sin(u), then dx = (√12)cos(u) du .

    Then [itex]\displaystyle \sqrt{12-x^2}=\sqrt{12-12\sin^2(u)}\,, [/itex] etc.
     
  4. Oct 21, 2011 #3

    Mark44

    Staff: Mentor

    As SammyS says, a trig substitution is the way to go here. Since I don't try to remember which trig substitution should be applied, I just sketch a right triangle, labeling the sides and hypotenuse according to the expression inside the radical.

    The expression 12 - x2 suggests that the hypotenuse is of length √12, and I can label the altitude as x, and the angle opposite the altitude as θ.

    From this I get sin(θ) = x/√12, or x = √12*sin(θ), which is essentially the same as what SammyS is recommending.
     
  5. Oct 22, 2011 #4
    Thanks Sammy and Mark,

    Will give it a go. Also, I think Mark's the first person I've encountered who's actually given a methodology for determining an appropriate substitution. Thanks :)

    Regards,

    Chris
     
  6. Oct 22, 2011 #5
    Well...

    It would seem I have fried my brain with the late nights spent trying to work this out. So far I've got;

    ∫cos3u.du

    and once again, I know the answer (=1/3*sin3u +sinu), but for the life of me I can't figure out how to get there. Is having the function in the form cos3u the right way to go? I swear I'm overlooking something blindingly obvious here.

    Regards,

    Chris
     
  7. Oct 22, 2011 #6
    OK,

    Well now I'm trying to use the trig identity cos3x = 4cos3x -3cosx

    Sound like the right way to go?

    Regards,

    Chris
     
  8. Oct 22, 2011 #7

    SammyS

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    That should work fine, but aren't you integrating cos2(u).
     
  9. Oct 22, 2011 #8
    Well now I'm not to sure. After substituting in,

    x = √12*sin(u)

    I got,

    dx = √12*cos(u)du

    Putting both of these back into the integral I then had,

    ∫√(12 -12*sin2(u))*√12*cos(u)du

    =∫√(12(1 -sin2(u))*√12*cos(u)du

    using the trig identity, cos2(u) +sin2(u) = 1

    I then realised whilst writing this that I didn't take the square root of the first cos term. Need sleep.....

    Thanks Sammy

    Regards,

    Chris
     
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