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    https://5c59854d0ccd29d489c9e5e689a8...%20Edexcel.pdf

    Q7c I cant see why you need to multiply cosec(t) by the derivative of cos(2t)
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    (Original post by Sniperdon227)
    https://5c59854d0ccd29d489c9e5e689a8...%20Edexcel.pdf

    Q7c I cant see why you need to multiply cosec(t) by the derivative of cos(2t)
    you want the integral of y with respect to x, not to t.
    thus dx/dt = -2sin2t -> dx = -2 sin 2t dt.
    Hence we require the integral of cosec t * -2 * 2 sin t cos t dt, i.e. integral of -4 cos t dt (since cosec t and sin t cancel out), so k = -4.
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    (Original post by Sniperdon227)
    https://5c59854d0ccd29d489c9e5e689a8...%20Edexcel.pdf

    Q7c I cant see why you need to multiply cosec(t) by the derivative of cos(2t)
    Because you integrate \int y \, \mathrm{d}x in cartesian which is the same as = \int y \, \frac{\mathrm{d}x}{dt} \times \mathrm{d}t in parametric. So to get the \mathrm{d}t as required in the answer you need to compensate by multiplying by the derivative of x with respect to t. It is not true that \mathrm{d}x = \mathrm{d}t.
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    Thanks i get it now, learnt something new
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    (Original post by Sniperdon227)
    Thanks i get it now, learnt something new
    Great! No problem.
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    (Original post by HapaxOromenon3)
    you want the integral of y with respect to x, not to t.
    thus dx/dt = -2sin2t -> dx = -2 sin 2t dt.
    Hence we require the integral of cosec t * -2 * 2 sin t cos t dt, i.e. integral of -4 cos t dt (since cosec t and sin t cancel out), so k = -4.
    Isnt k=4 because they swapped the limits around?
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    (Original post by metrize)
    Isnt k=4 because they swapped the limits around?
    MS:
    correct the limits switch, only because of the minus sign and because it wants you to
    https://5c59854d0ccd29d489c9e5e689a8...%20Edexcel.pdf
 
 
 
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