How to integrate 1/(sqrt (4 - y^2)) dy?

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  • s_h_mc
    Lv 4
    1 decade ago
    Favorite Answer

    Note that the derivative of arcsin(x) = 1/sqrt(1-x^2).

    So, if you let y = 2u, then y^2 = 4u^2, dy = 2 du.

    Our integrand becomes

    2*(1/sqrt(4 - 4u^2)) du

    Factor out 4 in the square root and then take the square root, you get

    2*(1/(2sqrt(1-u^2))) du

    The 2s in the numerator and denominator will reduce leaving

    1/(sqrt(1 - u^2)) du. The integral is arcsin(u) + 1.

    Back-substituting yields: arcsin(y/2) + C

  • 1 decade ago

    sqrt (4 - y^2) = sqrt(4(1 - (y^2)/4)) = 2sqrt(1-(y/2)^2)

    let (y/2) = sin(x)

    dy = cos(x) dx

    2sqrt(1-(y/2)^2) = 2sqrt(1-(sin(x))^2) = 2cos(x)

    int( 1/(sqrt (4 - y^2)) )dy =

    int( 1/2cos(x) )cos(x) dx = int(1/2)dx = x/2 + C

    x = arcsin(y/2), so int( 1/(sqrt (4 - y^2)) )dy = arcsin(y/2)/2 + C

  • cidyah
    Lv 7
    1 decade ago

    ∫ dy/ sqrt(2^2 -y^2)= arcsin(y/2) +C

    Note:

    Use the fact that ∫ dx/ sqrt(a^2-x^2) = arcsin(x/a)+C

  • 7 years ago

    int sqrt(y^4-1)

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  • Anonymous
    1 decade ago

    mathway.com

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