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Thread: Integration Problem

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    Member Nelly's Avatar
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    Integration Problem

    I have a Integration problem for all ya'll. Need description on how to do ASAP. Thank You:

    INT: dx/[x(x^2+a^2)^1/2]

    one on x times the square root of x squared plus a squared.
    "You're theory of a donut shaped universe is intriguing Homer, I may have to steal it" - Stephen Hawking

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    Re: Integration Problem

    Originally posted by Nelly
    I have a Integration problem for all ya'll. Need description on how to do ASAP. Thank You:

    INT: dx/[x(x^2+a^2)^1/2]

    one on x times the square root of x squared plus a squared.
    Using (d/dx) f(x)^1/2 = [2*f(x)^-1/2]/f'(x)
    The Integral of: dx/[x(x^2+a^2)^1/2]
    is (x^2 +a^2)^1/2

    the square root of x squared plus a squared

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    Member Nelly's Avatar
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    Huh? How does that work. I eventually got it. I used u^2=x^2+a^2. Then used implicit differentiation to find dx = u/x dx.

    If anyone has any integration problems, put them all up here:
    "You're theory of a donut shaped universe is intriguing Homer, I may have to steal it" - Stephen Hawking

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    oh...oops...i just realised i made a mistake
    sorry

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    Gender: MALE!!! -=«MÄLÅÇhïtÊ»=-'s Avatar
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    It takes too long to type out the working so i'll juz guide you

    You know that 1+(tanA)^2=(secA)^2

    So let x = atanA

    You will simplify it down to 1/a (Int.) dA/asinA

    Then you continue by letting t = tan(A/2)
    -= im a guy!!!!!!!!!!! =-

  6. #6
    Member Nelly's Avatar
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    So u saying use the t-formula??
    "You're theory of a donut shaped universe is intriguing Homer, I may have to steal it" - Stephen Hawking

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    of course

    then when reforming your integral, you'll need to use double angle formula to change the [cos(A/2)]^2

    It'll easily simplify b4 you integrate
    -= im a guy!!!!!!!!!!! =-

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