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Math Help - Integration

  1. #1
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    Integration

    Hi!

    Could some one please remind me how to integrate 1 / (square root of (9-4x^2)?

    Many thanks
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  2. #2
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    Quote Originally Posted by Natasha1 View Post
    Hi!

    Could some one please remind me how to integrate 1 / (square root of (9-4x^2)?

    Many thanks
    Int(1/sqrt(9-4x^2) dx) = Int(1/2 * 1/sqrt(9/4 - x^2) dx) (Where I have factored a sqrt(4) from the denominator.)

    Note that 9/4 = (3/2)^2 is a perfect square. Now make the trig substitution:
    x = 3/2 * sin(y) ==> y = asn(2x/3)
    dx = 3/2 * cos(y) dy

    Int(1/sqrt(9-4x^2) dx) = 1/2 * Int(1/sqrt(9/4 - 9/4sin^2(y)) 3/2*cos(y)dy)

    = 1/2 * 3/2 * Int( 1/(3/2) * 1/sqrt(1 - sin^2(y)) cos(y)dy) (I have factored a sqrt(9/4) from the denominator.)

    = 1/2 * 3/2 * 2/3 * Int(1/cos(y) * cos(y)dy)

    = 1/2 * Int(dy)

    = 1/2 * y + C

    = 1/2 * asn(2x/3) + C.

    -Dan
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Natasha1 View Post
    Hi!

    Could some one please remind me how to integrate 1 / (square root of (9-4x^2)?

    Many thanks
    int 1/sqrt(9-4x^2) dx = (1/3) int 1/sqrt(1-(4/9)x^2) dx

    let sin(u) = (2/3)x,

    then cos(u) du/dx = 2/3, so our integral becomes:

    (1/3) int 1/sqrt(1-(4/9)x^2) dx

    .........................= (1/3) int 1/sqrt(1 - sin^2(u)) (3/2) cos(u) du

    .........................= (1/2) int 1/cos(u) cos(u) du

    .........................= (1/2) int du = u/2 = (1/2) arcsin( 2x/3) + C

    RonL
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  4. #4
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    Quote Originally Posted by Natasha1 View Post
    Hi!

    Could some one please remind me how to integrate 1 / (square root of (9-4x^2)?

    Many thanks
    I in fact have the simplest method

    Use the linear substitution u=2x
    Thus,
    INTEGRAL 1/sqrt(9-4x^2) = 1/2*INTEGRAL 1/sqrt(9-u^2)
    Factor the 9 (becomes 3 outside the radical),
    1/6*INTEGRAL 1/sqrt[1-(u/3)^2]
    Use linear substitution t=u/3,
    3/6*INTEGRAL 1/sqrt[1-t^2] = 1/2*INTEGRAL 1/sqrt(1-t^2)
    This is a tabular integral (you must have it burned in your memory),
    Thus,
    1/2*asin(t)+C
    t=u/3
    Thus,
    1/2*asin(u/3)+C
    u=2x
    Thus,
    1/2*asin(2x/3)+C
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