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Trigonometric Substitution 11x^2dx/(25-x^2)^(3/2)

  1. Jun 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Integral (11x^2)/(25-x^2)^(3/2) dx from 0 to (5*sqrt(3))/2


    2. Relevant equations

    sin^2(θ) = 1 - cos^2(θ)


    3. The attempt at a solution

    1. Factor out 11 from integral for simplicity.

    11 * integral (x^2)/(25-x^2)^(3/2)

    2. Re-write denominator of integral to look similar to 1-cos^2(θ)

    11 * integral (x^2)/(25(1-(1/25)x^2)))^(3/2)dx

    3. Equate cos^2(θ) and (1/25)x^2

    cos(θ) = (1/5)x
    θ = arccos(1/5*x)
    x = 5cos(θ)
    dx = -5sin(θ)dθ

    4. Substitute cos^2(θ) = (1/25)x^2 into integral, Substitute -5sin(θ)dθ = dx

    11*(-5) * integral (x^2)(sin(θ)/(25(1-cos^2(θ)))^(3/2)

    5. Substitute sin^2 for (1-cos^2)

    -55 * integral (x^2)(sin(θ)/(25(sin^2(θ))^(3/2)

    No idea what to do from here.

    I have a feeling my approach in general is totally off. Any advice would be greatly appreciated.
     
  2. jcsd
  3. Jun 13, 2013 #2
    If x^2/25=cos^2(θ), what is x^2? Substitute that in your final expression.
     
  4. Jun 13, 2013 #3
    Hi Pranav-Arora. I'll try...


    -55 * integral (25cos^2(θ))(sin(θ)/(25(sin^2(θ))^(3/2)

    Not sure how to proceed from there. I gotta go to classes now, but I'll be back to think more about the problem shortly.

    I appreciate your help.
     
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