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¡Ò dy/(rad£¨4-y^2))=

  • Calculus -

    Evaluate the indefinite integral

  • Calculus -

    look into trig substitutions

    let y = 2sinu

    then 4 - y^2 = 4 - 4sin^2 u = 4 cos^2 u
    and √(4 - y^2) = 2cos u
    dy = 2 cosu du

    and your integral is just

    1/(2 cosu) * 2 cosu du = 1/2 du

    integral(1/2) du = u/2 = (1/2) arcsin(y/2) + C

  • Calculus - oops -

    oops - too many 1/2's there. integral(du) = u = arcsin(y/2)+C

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