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Math Help - Test

  1. #1
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    Joined
    Nov 2008
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    92

    Test

    Hello

    In this post integralof is the integration sign and th is theta.

    1) Integrate x(2x^2-3)^5

    Let u = x, du = 1
    dv = (2x^2-3)^5, v = (2x^2-3)^6/24x

    I = x(2x^2-3)^6/24x - integralof(2x^2-3)^6/24x dx

    I do not know how to carry this on. The answer is (2x^2-3)^6/24 + k

    2) Substituting a trigonometric function for x, integrate x^2/root(4-x^2)

    Let  x = 2sin(th)

    I = integrate(4sin^2(th))/4cos^2(th) * 2cos(th) d(th))
    The terms cancel
    I = integralof(4sin^2(th) d(th))
    I = integralof(2 - 2cos(2th) d(th))
    I = 2th - sin(2th) + C

    th = sin^-1(x/2)

    :. I = 2sin^-1(x/2) - sin(2th)

    How do I manipulate sin(2th) in terms of x?

    The answer is 2sin^-1(x/2) - (x/2)root(4-x^2) + k

    Thank you
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  2. #2
    Member
    Joined
    Nov 2008
    Posts
    92
    For 1, it asks for "substitution where necessary". How would it be done using this method?

    Thank you
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