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Math Help - integration - urgent help please

  1. #1
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    Question integration - urgent help please

    the integral from 0 to pi/2 of acost *bsint *sqrt(a^2*sin^2 t + b^2*cos^2 t)dt

    Thank you very much.
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  2. #2
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    hi turkeywilliam,

    Thank you very much for your reply.

    why ? Could you please explain to me for this link. thank you .


    also, i have the final answer, given by my teacher, is [ab(a^2 +ab+b^2) ]/3(a+b).
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  3. #3
    Senior Member tukeywilliams's Avatar
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     ab \int_{0}^{\pi/2}  \sin t \cos t \sqrt{a^{2} \sin^{2}t + b^{2} \cos^{2} t} \ dt

    Let  x = a \cos t and  y = b \sin t (elliptic integral).
    Last edited by tukeywilliams; October 28th 2007 at 04:45 AM.
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  4. #4
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    INT.(0-->pi/2)[(a*sinT)(b*cosT)sqrt[(a^2)sin^2(T) +(b^2)cos^2(T)]]dT

    = INT.(0-->pi/2)[sqrt[(a^2)sin^2(T) +(b^2)cos^2(T)]*[ab*sinTcosT dT] -----(i)

    (a^2)sin^2(T) +(b^2)cos^2(T)
    = (a^2)(1 -cos^2(T)) +(b^2)cos^2(T)
    = a^2 -(a^2)cos^2(T) +(b^2)cos^2(T)
    = a^2 -(a^2 -b^2)cos^2(T)

    And the derivative of that is
    d/dt [a^2 -(a^2 -b^2)cos^2(T)]
    = 0 -(a^2 -b^2)*(2cosT)(-sinT dT)
    = 2(a^2 -b^2)sinTcosT dT

    Use that in (i),
    INT.(0-->pi/2)[sqrt{(a^2)sin^2(T) +(b^2)cos^2(T)}]*[ab*sinTcosT dT]

    = INT.(0-->pi/2)[sqrt{(a^2)sin^2(T) +(b^2)cos^2(T)}]*[ab*sinTcosT dT]*[{2(a^2 -b^2)/ab} /

    {2(a^2 -b^2)/ab}]

    = [ab/ 2(a^2 -b^2)]*INT.(0-->pi/2)[sqrt{(a^2)sin^2(T) +(b^2)cos^2(T)}]*[2(a^2 -b^2)sinTcosT

    dT]

    That is in the form
    INT.[u^(1/2)]du = (2/3)U^(3/2) +C

    So,
    = [ab/ 2(a^2 -b^2)](2/3){(a^2)sin^2(T) +(b^2)cos^2(T)}*(3/2) | (0-->pi/2)
    = [ab/ 3(a^2 -b^2)]{(a^2)sin^2(T) +(b^2)cos^2(T)}*(3/2) | (0-->pi/2)

    = [ab/ 3(a^2 -b^2)]*[{(a^2)sin^2(pi/2) +(b^2)cos^2(pi/2)}^(3/2) -{(a^2)sin^2(0)

    +(b^2)cos^2(0)}^(3/2)}]

    = [ab/ 3(a^2 -b^2)]*[{(a^2)(1) +(b^2)(0)}^(3/2) -{(a^2)(0) +(b^2)(1)}^(3/2)}]

    = [ab/ 3(a^2 -b^2)]*[(a^2)^(3/2) -(b^2)^(3/2)]

    = [ab/ 3(a^2 -b^2)]*[a^3 -b^3]

    = (1/3)[ab(a^3 -b^3) / (a^2 -b^2)]

    = (1/3)[ab(a-b)(a^2 +ab +b^2) / (a-b)(a+b)]

    = (1/3)[ab(a^2 +ab +b^2) / (a+b)] ---------------------answer.
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