Integration Question Solomon Help

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Q7c I cant see why you need to multiply cosec(t) by the derivative of cos(2t)

Reply 1

HapaxOromenon3

3

Original post by Sniperdon227

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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.

Reply 2

Zacken

22

Original post by Sniperdon227

https://5c59854d0ccd29d489c9e5e689a8bbadf49aa0f0.googledrive.com/host/0B1ZiqBksUHNYREhxMHhfam1IQm8/for-Edexcel/Solomon%20A%20QP%20-%20C4%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

.

Reply 3

Sniperdon227

OP

5

Thanks i get it now, learnt something new

Reply 4

Zacken

22

Original post by Sniperdon227

Thanks i get it now, learnt something new

Great! No problem. :smile:

Reply 5

metrize

18

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.