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Calc

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I have to find the integral of 1/(sq. rt. of (4x-x²)) dx. I know I have to complete the square of the denominator, but im confused on how that process works when im only given 4x-x². Could you demonstrate how to complete the square for that equation? Thank you so much!

  • Calc - ,

    4x - x^2
    -(x^2 - 4x)
    x^2 - 4x = 0
    x^2 - 4x + 4 = 0 + 4
    (x - 2)^2 = 4
    (x - 2)^2 - 4
    -(x - 2)^2 + 4
    4 - (x - 2)^2

    | = integral sign
    | 1/(sqrt(4 - (x - 2)^2) dx

    u = ( x - 2)
    du = dx

    | 1/(sqrt(4 - u^2)) du
    = arcsin u/2 + C
    = arcsin (x - 2)/2 + C

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