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Integral x^3sqrt(4x^2 -x^4) dx

  1. Oct 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Integral x^3sqrt(4x^2 -x^4) dx

    2. Relevant equations

    Maybe something from a table

    3. The attempt at a solution

    I pulled the x^2 our from under the square root.

    ∫ x^4 (sqrt(4-x^2)) dx

    I'm not sure how I can do this one.
    Maybe it can fit the forum ∫ usinu du ?
     
  2. jcsd
  3. Oct 5, 2013 #2
    Yes you started right. Now make the substitution x = 2cos u.
     
  4. Oct 6, 2013 #3
    I sub in 2sinθ instead. Same deal.

    ∫x^4 √(4-x^2) dx

    ∫ (2sinθ)^4 √(1-sinθ^2)cosθ dθ I did 2cosθ dθ = dx

    (2^4)(4) ∫ sinθ^4cosθ^2 dθ

    There is a formula in my book which has ∫sinu^nscosu^m du = sinu^(n+1)cosu^(m-1)/(m+ n)
    + (m-1)/(n+m)∫ sinu^ncosu^(m-2)du

    When I pop my stuff in I got

    I = (sinθ^5cosθ)/6 + (1/6) ∫ sinθ^4 dθ ...for ∫ sinθ^4 dθ I used a reduction formula and I got (-1/4)sinθ^3cosθ + (3/4)∫ sinθ^2 dθ So all together I ended up with

    (64)[ (sinθ^5cosθ)/6 -(1/24)sinθ ^3cosθ + (3/8)θ -(3/8)sinθcosθ]

    Then I put it back in terms of x I got (32/3)(x/2)^5(√(4-x^2)/2 -(8/3)(x/2)^3(√(4-x^2)/2 +24arcsin(x/2) -24(x/2) (√(4-x^2)/2

    I have to evaluate it from [0,2] I should come out to 2PI like on my calcualtor but it doesn't ...please help me sort this out where did I go wrong?
     
  5. Oct 6, 2013 #4
    Been away all day, will get back to you soon.
     
  6. Oct 6, 2013 #5
    ##\int x^3 \sqrt{4x^2 - x^4}dx = \int x^4 \sqrt{4-x^2}dx \hspace{100px} ##(1)

    Let x = 2sinu and dx = cosu du so (1) becomes

    ##\int (2sinu)^4 \sqrt{4 - 4sin^2u}(2cosu) du \hspace {100px}##. (2)

    Gathering up factors of 2 (and please fix if I get it wrong) (2) becomes

    ## 2^6\int (sin^4u)(cos^2u)du = 2^6\int (sin^4u)(1 - sin^2u)du = 2^6\int (sin^4u - sin^6u)du## (or you can do it in cosines if you prefer).

    There are a number of ways to integrate this, but one approach is to deflate the powers of sin by using the half angle formulas:

    ##sin^2u = \frac{1-cos(2u)}{2}\text { and } cos^2u = \frac{1+cos(2u)}{2}. ## You will have to use this several times but eventually you will get down to something you can integrate. Along the way, you will shed the various powers of 2 that accumulated.

    When you finally get down to the definite integral, if you are not getting the right answer, let me know. We'll go over it very carefully.

    If this problem was given to you in a first year calc course, it does seem a bit much. I have played around with it to see if I could find an easy solution, but so far I haven't. If your ##x^3## were ##x^{3/2}## it would be much easier; maybe you should double check, just in case that was it.
     
    Last edited: Oct 7, 2013
  7. Oct 8, 2013 #6
    I solved this thanks for your help
     
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