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sorry, yet another question watch

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    by using the fact that sinx >= (2/pi)x for 0<= x <= pi/2, show that:

    integral between pi/2 and 0 of: (x^2)/(1+sin^2) < (pi^3 / 8)(1-pi/4)
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    i dont understand that last step:


    < pi^3/8 < (pi^3 / 8)(1-pi/4) QED

    that's surely wrong....pi^3/8 > (pi^3 / 8)(1-pi/4)
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    Sorry, I integrated the wrong way round. Initially I integrated between 0 and pi/2 but I just read in the original question that you asked for between pi/2 and 0. In that case:

    Start off by saying:

    sinx >= (2x/pi)
    (sinx)^2 >= (2x/pi)^2
    1 + (sinx)^2 > (2x/pi)^2

    1/[1 + (sinx)^2] < 1/(2x/pi)^2
    x^2/[1 + (sinx)^2] < pi^2/4

    Integrate the RHS:

    < pi^2.x/4

    Putting in the limits (pi/2 ... 0):

    < -[pi^3]/8 < (pi^3 / 8)(1-pi/4) QED


    Euclid.
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    no no...you were right the first time....the upper bound of the integral is pi/2, the lower bound is 0

    so it's still not right!
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    (Original post by Willa)
    no no...you were right the first time....the upper bound of the integral is pi/2, the lower bound is 0

    so it's still not right!
    Why dont you try to find the integral of x^2/(1+ (2x/pi)^2) from 0 ->pi/2. It maybe work
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    (Original post by Willa)
    no no...you were right the first time....the upper bound of the integral is pi/2, the lower bound is 0

    so it's still not right!
    Arrghh! Ok this one is definately right!:

    sinx >= (2x/pi)
    (sinx)^2 >= (2x/pi)^2
    1 + (sinx)^2 >= 1 + (2x/pi)^2

    1/[1 + (sinx)^2] <= 1/[1+ (2x/pi)^2]
    x^2/[1 + (sinx)^2] <= [x^2]/[1+ (2x/pi)^2]

    x^2/[1 + (sinx)^2] <= (pi^2).(x^2)/(pi^2 + 4x^2)

    Use the substitution u = 2x/pi, and then integrating the RHS will get:

    (x.pi^2)/4 - (arctan(2x/pi).pi^3)/8

    Putting in the limits (0 ... pi/2):

    <= (pi^3)/8 - (pi^4)/32
    <= [(pi^3)/8](1 - pi/4) ..........

    Euclid.
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    (Original post by Euclid)
    Arrghh! Ok this one is definately right!:

    sinx >= (2x/pi)
    (sinx)^2 >= (2x/pi)^2
    1 + (sinx)^2 >= 1 + (2x/pi)^2

    1/[1 + (sinx)^2] <= 1/[1+ (2x/pi)^2]
    x^2/[1 + (sinx)^2] <= [x^2]/[1+ (2x/pi)^2]

    x^2/[1 + (sinx)^2] <= (pi^2).(x^2)/(pi^2 + 4x^2)

    Use the substitution u = 2x/pi, and then integrating the RHS will get:

    (x.pi^2)/4 - (arctan(2x/pi).pi^3)/8

    Putting in the limits (0 ... pi/2):

    <= (pi^3)/8 - (pi^4)/32
    <= [(pi^3)/8](1 - pi/4) ..........

    Euclid.
    Yeah I know it works . But I'm lazy. Why didn't u just give him a clue? :cool:
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    (Original post by BCHL85)
    Yeah I know it works . But I'm lazy. Why didn't u just give him a clue? :cool:
    Ah well! I think the first one didn't work because of the loss of accuracy when adding 1
 
 
 
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