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Integrating with a given substitution

  1. Nov 4, 2015 #1
    1. The problem statement, all variables and given/known data
    Using the substitution x = 2sinθ, show that
    [tex] \int \sqrt{4 - x^2} dx = Ax\sqrt{4 - x^2} + B ⋅ arcsin(\frac{x}{2}) + C [/tex]
    whee A and B are constants whose values you are required to find.
    2. Relevant equations


    3. The attempt at a solution
    x = 2sinθ
    [itex] \frac{dx}{dθ} = 2cosθ [/itex]
    dx = 2cosθ ⋅ dθ
    [itex] \int \sqrt{4 - x^2} dx = \int \sqrt{4 - 4sin^2θ} ⋅ 2cosθ dθ [/itex]
    Edit: (Hopefully) corrected my work after this point
    [itex] 2\int \sqrt{4(1 - sin^2θ)} ⋅ cosθ dθ [/itex]
    [itex] 2\int 2\sqrt{1 - sin^2θ} ⋅ cosθ dθ \rightarrow 1 - sin^2θ = cos^2θ [/itex]
    [itex] 4\int \sqrt{cos^2θ} ⋅ cosθ dθ [/itex]
    [itex] 4 \int cosθ ⋅ cosθ ⋅ dθ [/itex]
    [itex] 4 \int cos^2θ ⋅ dθ [/itex]
    [itex] 4 \int \frac{1 + cos2θ}{2} ⋅ dθ = 2 \int 1 + cos2θ ⋅ dθ [/itex]
    2(θ + 2sin2θ)
    x = 2sinθ ∴ θ = arcsin[itex]\frac{x}{2}[/itex]
    2arcsin[itex]\frac{x}{2}[/itex] + 4sin2θ


    I'm getting closer at this point, but can't figure out how to transform the 4sin2θ back into something to do with x's. If I substituted 2sinθcosθ in for sin2θ then I would be stuck with a cosine in the final equation that I don't want, nor would that get me any closer to a square root shown in the final answer. Have I messed up in my work up to this point again, or is there something I'm not looking at? Thanks!
     
    Last edited: Nov 4, 2015
  2. jcsd
  3. Nov 4, 2015 #2

    blue_leaf77

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    I don't see where that comes from, is second line supposed to be the second term of the expression to the right of the equal sign in the first line?
     
  4. Nov 4, 2015 #3
    For whatever reason, during my work I felt integration of one part of one term was possible. I'm redoing my work now, thanks for pointing out my mistake!
     
  5. Nov 4, 2015 #4
    I've redone all my work, still can't seem to get the final bit, but I'm getting closer.
     
  6. Nov 4, 2015 #5

    blue_leaf77

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    That's the right direction. Remember that ##\cos \theta = \sqrt{1-\sin^2 \theta}##, from this point substitute ##x/2 = \sin \theta##.
     
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