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    • Thread Starter
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    if you have a p3 book its ex 4e q 47

    use the sub x=3sint to show that

    intergration limits (3,0) x^2(9 - x^2)^0.5 dx = 81(pie)/16

    using theories below and including p3 level please
    • PS Reviewer
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    PS Reviewer
    I remember this one! Basically, you sub x=3sint into the equation, and then mess it around a bit and it will make something very nice to integrate. I would do it for you but I've had loads of codeine
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    x = 3sint
    dx = 3cost dt

    Integral becomes:
    9sin²t.(9-9sin²t).(3cost) dt
    9sin²t.3cost.3cost dt [using 1-sin²t=cos²t]
    81sin²tcos²t dt
    20.25(4sin²tcos²t) dt
    20.25(sin2t)^2 dt [using sin2t=2sintcost]
    20.25(1-cos4t)/2 dt [using cos4t=1-2(sin2t)^2]
    10.125 [t - (sin4t)/4]
    Limits become:
    3 -> pi/2
    0 -> 0
    So integral becomes:
    10.125[(pi/2 - 0) - (0)] = 10.125pi/2 = 81pi/16

    Hmm.. that was long-winded. I think there's a better way to do it, but I'm just not seeing it right now.
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    thanks dvs but anyone got a better method
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    (Original post by BLUREMI)
    Use the sub. x = 3sint to show that:
    Int. [limits (3,0)] x^2(9 - x^2)^(1/2) dx = (81pi)/16
    Let x = 3sint --> dx/dt = 3cost --> dx = 3cost dt

    Hence: Int. x^2(9 - x^2)^(1/2) dx
    = Int. 9sin^2t(9 - 9sin^2t)^(1/2) . 3cost dt
    = Int. 9sin^2t[9(1 - sin^2t)]^(1/2) . 3cost dt
    = Int. 9sin^2t(9cos^2t)^(1/2) . 3cost dt
    = 81 Int. sin^2tcos^2t dt
    = (81/4) Int. 4sin^2tcos^2t dt
    = (81/4) Int. [2sintcost]^2 dt
    = (81/4) Int. sin^2(2t) dt
    = (81/4) Int. 1/2(1 - cos4t) dt
    = (81/8) [t - (sin4t)/4] + k
    = 81t/8 - 81(sin4t)/32 + k
    = [324t - 81(sin4t)]/32 + k

    To find the new limits:

    When x = 3: --> 3 = 3sint --> sint = 1 --> t = Pi/2
    When x = 0: 0 = 3sint --> sint = 0 --> t = 0

    Hence: Int. [limits (3,0)] x^2(9 - x^2)^(1/2) dx
    = {324t - 81(sin4t)]/32} [limits Pi/2 and 0]
    = (162Pi)/32
    = (81Pi)/16

    Nima
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    That's the same way I did it. So I take it there's no simpler way?
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    (Original post by Nima)
    Sorry lol, I didn't copy your post, honest. I saw the thread starter's post and tried to do it in the simplest way I could. I think it is the simplest way - Originally I was going to do it by parts until I spotted that 4sin^2tcos^2t is useful...

    Nima
    lol, I saw you many times did like that. I thought you just try to do whatever you see, don't you?
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    (Original post by Nima)
    Sorry lol, I didn't copy your post, honest.
    I didn't mean it like that. I'm sorry if it sounded as if I did.
    • Thread Starter
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    do u think questions could be that hard in the p3 exam?
 
 
 
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