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Math Help - Trig Substitution

  1. #1
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    Trig Substitution

    These are 2 problems that I'm stumped on and can't get...we are supposed to evaluate the integral using trig substitution.

    x^2/(3+4x-4x^2)

    (x^2+1)/(x^2-2x+2)
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  2. #2
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    Quote Originally Posted by sfgiants13 View Post
    These are 2 problems that I'm stumped on and can't get...we are supposed to evaluate the integral using trig substitution.

    x^2/(3+4x-4x^2)

    [snip]
    \frac{x^2}{3 + 4x - 4x^2} = -\frac{1}{4} + \frac{x + \frac{3}{4}}{-4x^2 + 4x + 3}

    A trig substitution is not needed here because the quadratic is not irreducible. use a partial fraction decomposition
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    Quote Originally Posted by sfgiants13 View Post
    [snip]
    (x^2+1)/(x^2-2x+2)
    \frac{x^2 + 1}{x^2 - 2x + 2} = \frac{(x^2 - 2x + 2) + 2x - 1}{x^2 - 2x + 2} = 1 + \frac{2x - 1}{x^2 - 2x + 2}


    = 1 + \frac{(2x - 2) + 1}{x^2 - 2x + 2} = 1 + \frac{2x - 2}{x^2 - 2x + 2} + \frac{1}{x^2 - 2x + 2}


    1 + \frac{2x - 2}{x^2 - 2x + 2} + \frac{1}{(x - 1)^2 + 1}.


    The first two terms are trivial to integrate. To integrate the last term make the substitution x-1 = tan \theta (or just recognise the standard form).
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