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# Integration Question Solomon Help

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

Q7c I cant see why you need to multiply cosec(t) by the derivative of cos(2t)
2. (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.
3. (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 in cartesian which is the same as in parametric. So to get the as required in the answer you need to compensate by multiplying by the derivative of with respect to . It is not true that .
4. Thanks i get it now, learnt something new
5. (Original post by Sniperdon227)
Thanks i get it now, learnt something new
Great! No problem.
6. (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?
7. (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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