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Math Help - trig integral

  1. #1
    Member jacs's Avatar
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    Unhappy trig integral

    Hi, i just can't seem to get this one out, this is what i have so far (see attachment)

    any help would be greatly appreciated.

    the solution is 1/9


    thanks

    jacs
    Attached Thumbnails Attached Thumbnails trig integral-integral.gif  
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  2. #2
    MHF Contributor
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    Quote Originally Posted by jacs
    Hi, i just can't seem to get this one out, this is what i have so far (see attachment)

    any help would be greatly appreciated.

    the solution is 1/9


    thanks

    jacs
    Here is one way.

    INT.(0-->3/sqrt(2))[(9 -x^2)^(-3/2)]dx

    Let x = 3sinA ----------***
    So,

    dx = 3cosA dA ----**

    The new boundaries, for dA,
    When x = 0,
    0 = 3sinA
    0 = sinA
    A = arcsin(0) = 0 ----**
    When x = 3/sqrt(2),
    3/sqrt(2) = 3sinA
    1/sqrt(2) = sinA
    A = arcsin(1/sqrt(2)) = pi/4 ---**

    (9 -x^2)^(-3/2)
    = (9 -9sin^2(A))^(-3/2)
    = 1 / [9(1-sin^2(A)]^3/2
    = 1 / 27*(cos^2(A))^(3/2)
    = 1 / 27cos^3(A)

    So,
    INT.(0-->3/sqrt(2))[(9 -x^2)^(-3/2)]dx
    = INT.(0-->pi/4)[1 / 27cos^3(A)]*3cosA dA
    = (3/27)INT.(0-->pi/4)[1 / cos^3(A)]*cosA dA
    = (1/9)INT.(0-->pi/4)[1 / cos^2(A)]dA
    = (1/9)INT.(0-->pi/4)[sec^2(A)]dA
    = (1/9)[tanA](0-->pi/4)
    = (1/9)[tan(pi/4) -tan(0)]
    = (1/9)[1 -0]
    = 1/9 ---------------------answer.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by jacs
    Hi, i just can't seem to get this one out, this is what i have so far (see attachment)

    any help would be greatly appreciated.

    the solution is 1/9


    thanks

    jacs
    You're almost there. Note that \frac{d}{d \theta}cot\theta = -csc^2 \theta.

    -Dan
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  4. #4
    Member jacs's Avatar
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    Quote Originally Posted by ticbol
    Here is one way.

    INT.(0-->3/sqrt(2))[(9 -x^2)^(-3/2)]dx

    Let x = 3sinA ----------***
    So,

    dx = 3cosA dA ----**

    The new boundaries, for dA,
    When x = 0,
    0 = 3sinA
    0 = sinA
    A = arcsin(0) = 0 ----**
    When x = 3/sqrt(2),
    3/sqrt(2) = 3sinA
    1/sqrt(2) = sinA
    A = arcsin(1/sqrt(2)) = pi/4 ---**

    (9 -x^2)^(-3/2)
    = (9 -9sin^2(A))^(-3/2)
    = 1 / [9(1-sin^2(A)]^3/2
    = 1 / 27*(cos^2(A))^(3/2)
    = 1 / 27cos^3(A)

    So,
    INT.(0-->3/sqrt(2))[(9 -x^2)^(-3/2)]dx
    = INT.(0-->pi/4)[1 / 27cos^3(A)]*3cosA dA
    = (3/27)INT.(0-->pi/4)[1 / cos^3(A)]*cosA dA
    = (1/9)INT.(0-->pi/4)[1 / cos^2(A)]dA
    = (1/9)INT.(0-->pi/4)[sec^2(A)]dA
    = (1/9)[tanA](0-->pi/4)
    = (1/9)[tan(pi/4) -tan(0)]
    = (1/9)[1 -0]
    = 1/9 ---------------------answer.

    thanks for that ticbol, but unfortunately, i am restricted to using the original substitution x = 3 cos A ggrrrrrr

    but many thanks anyway, alwasy most excellent to have an alternative

    jacs
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  5. #5
    Member jacs's Avatar
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    thanks Dan, i can see it soooo clearly now. I should have figured that out by myself. Now i can trot off and do the rest, thanks.


    jacs
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