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    x = cosθ
    dx = - sinθ dθ

    I(n) = - pi/20 cosnθ . sinθ . sinθ dθ
    = - pi/20 cosn sin²θ dθ
    = 0pi/2 cosn sin²θ dθ

    You want I(5).
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    Very sorry guys. I made a typo in the question. It's:

    In = ∫ xn.√(1-x)

    Heh. The reason I typed (1-x^2), was because I tried the cosθ substitution, and frustratingly got the (1-x^2) rather than (1-x). Must've confused me or something.
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    (Original post by Lucky Penny)
    god how did you do the first part??
    I would expect it was integration by parts. Differentiate x^{n} and integrate \sqrt{1-x}
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    (Original post by mockel)
    Very sorry Gaz. I made a typo in the question. It's:

    In = ∫ xn.√(1-x)
    Then you're going to have to use x=cos²θ with dx=-2sinθcosθ, I think.
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    Cool, I'll give it a go...
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    In the P5 exam, are we expected to know the t substitution and....for example what dt/dx is in this case...?
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    (Original post by lgs98jonee)
    In the P5 exam, are we expected to know the t substitution and....for example what dt/dx is in this case...?
    t = tan(x/2)?
    Well you can always derive dt/dx (very quick), along with the sin and cos ones (these are given, but also quick to derive)

    Oh, and the reduction one looks good now. Cheers.
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    btw dont know if u noticed but the t=tan(x/2) is in F/B on page6 at top
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    Has anyone done question 8 june 2004?! please help!
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    lol see other post! and stick to 1 post only dont spam
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    Which part do you need help with?
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    I am surprised about not being able to do this for sure.#

    differentiate y=x^sinhx

    is it dy/dx=coshx.sinhx.a^sinhx

    :/
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    hey i got stuck on either question 3 or 4 in the exam paper at the end of the book, its the one that involves y=mx+c... runs to room to get book
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    If that's from the sample exam at the end of the P5 book, then it should be y=x^x, not x^sinhx.
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    er which other post would that be syncman??

    the first part dvs..
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    hahahaha its impossible i spent hours trying it! back of P5 book.. should be y= x^x
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    (Original post by lgs98jonee)
    I am surprised about not being able to do this for sure.#

    differentiate y=x^sinhx

    is it dy/dx=coshx.sinhx.a^sinhx

    :/
    Is this from the exam style paper at the back of the P5 book? If so there's an error where x^{sinhx} should read x^{x}.
    If not:
    y=x^{sinhx}\\

lny=sinhxlnx\\

\frac{1}{y} \frac{dy}{dx}=\frac{sinhx}{x}+co  shxlnx\\

\frac{dy}{dx} = [ x^{sinhx} ][\frac{sinhx}{x}+coshxlnx]\\
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    (Original post by Lucky Penny)
    er which other post would that be syncman??

    the first part dvs..
    http://www.thestudentroom.co.uk/show...97&postcount=9
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    (Original post by Lucky Penny)
    er which other post would that be syncman??

    the first part dvs..
    xy = c²
    y + x(dy/dx) = 0
    dy/dx = -y/x

    So at P, dy/dx = -(c/t)/ct = -1/t². And hence the gradient of the normal is t², and it's equation is:
    y - c/t = t²(x - ct)
    y = t²x + c/t - ct³
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    (Original post by Lucky Penny)
    Has anyone done question 8 june 2004?! please help!
    8. a)
    P(ct,c/t)

    y=ct-1
    dy/dt = -ct-2

    x=ct
    dx/dt = c

    dy/dx = dy/dt . dt/dx
    =-c/(ct²)
    Tangent gradient = -1/t²
    hence normal = t²

    t - y1 = m(x-x1)
    y-c/t=t²(x-ct)
    y=t²x-ct3+c/t
 
 
 
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