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Integral of Sqrt of (49-4x^2) and Differentiating Answer

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\int\sqrt{49-4x^2}[/tex]
    I have done the integration to the best of my ability. My teacher requires that I differentiate the answer to get back to the original integrand. Can someone please check my work and help me along with the differentiation?

    2. Relevant equations
    U substitutions, and double angle formula ((sinx)^2=1/2 - (cosx)/2

    3. The attempt at a solution

    [tex]7\int\sqrt{1-(2x/7)^2}[/tex]

    cosu=2x/7
    -sinu du=2/7 dx
    u=arccos(2x/7)

    Integral Rewritten as:

    [tex]\int[/tex][tex]7\sqrt{1-cos^2}* -7(sinu)/2 du[/tex]
    [tex]7\sqrt{1-cos^2}= sinu[/tex] therefore:

    =[tex]-49/2\int(sin^2u)du[/tex]
    =[tex]-49/2\int(1-cos^2)/2 du[/tex]
    =[tex]-49x/4 + 49(sinu)/2 + C[/tex]
    =[tex]-49x/4 + 49(sin(arccos(2x/7)))[/tex])/4 + C[/tex]

    Differentiation:
    [tex]-49/4 + [49(cos(arccos(2x/7))]/(4 \sqrt{1-(2x/7)^2}) * 4/49[/tex]
    =[tex]-49/4 + (2x)/(7\sqrt{49-4x^2})[/tex]


    Does this make sense at all?
    How can I get to the original integrand given [tex]\int\sqrt{49-4x^2}[/tex]
     
    Last edited: Mar 17, 2009
  2. jcsd
  3. Mar 17, 2009 #2

    gabbagabbahey

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    Careful, the square root sign always returns the positive root

    [tex]\implies \sqrt{\sin^2(u)}=\left\{\begin{array}{lr}\sin(u), & \sin(u)\geq 0 \\ -\sin(u), & \sin(u)\leq 0\end{array}[/tex]

    For this reason, [itex]\cos(u)=\frac{2x}{7}[/itex] is a poor choice of substitution.

    Here, there is another error; [tex]\int \cos^2(u) du \neq -2\sin(u)[/tex]
     
  4. Mar 17, 2009 #3
    How do you suggest I solve it? I have literally spent days on this problem and I'm still lost :(
     
  5. Mar 17, 2009 #4

    gabbagabbahey

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    Try integration by parts instead.

    Edit there is a little trick necesary, so show me what you get after applying integration by parts once, and I'll help you from there
     
    Last edited: Mar 17, 2009
  6. Mar 17, 2009 #5
    I have tried Integration by Parts and it seems to never end.

    I separated the integrand into 2 fractions, by multiplying top and bottom by sqrt(49-4x^2) to get:
    49/ sqrt(49-4x^2) - 4x^2/(49-4x^2)
    The 1st fraction I can integrate easily, the second I used Integration by Parts so that:

    u=x^2
    thus du=2x

    dv=1/(1-(2x/7))
    v=7/2 * arcsin(2x/7)

    I kept doing IBP up to 5 times, and could not get an answer.
    According to the book the answer should be xsqrt(49-4x^2) + 49/2 arcsin(2x/7) + C
     
  7. Mar 17, 2009 #6

    gabbagabbahey

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    Hmmm... why not just use dv=dx and u=sqrt(49-4x^2)...what do you get after applying IBP once like this?
     
  8. Mar 17, 2009 #7

    gabbagabbahey

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    Good :smile:.... now comes the tricky business; let's call your original integral [itex]I[/itex]:

    [tex]I\equiv \int \sqrt{49-4x^2)dx[/tex]

    What happens if you add and subtract 49 to the numerator of integral of [4x^2/(Sqrt(49-4x^2)]?:wink:
     
  9. Mar 17, 2009 #8

    gabbagabbahey

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    Yes,

    [tex]\int\frac{49}{\sqrt{49-4x^2}}dx=\int\frac{49}{7\sqrt{1-\left(\frac{2x}{7}\right)^2}}dx[/tex]

    You forgot to divide by that 7
     
  10. Mar 17, 2009 #9

    gabbagabbahey

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  11. Mar 17, 2009 #10
    hum... sqrt(sin^2(x))=abs(sin(x)). in trig substitutions, as far as i know, we almost always take the triangle to be in the first quadrant, so i dont think we should be so particular about this.

    so sqrt(1-x^2)=cos(t) when x=sin(t) . here i prefer that x is vertical instead of horizontal.

    cos^2(x)=(1+cos(2t))/2 and the result should follow. i think this was what the op was looking for.
     
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