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1994 - Paper 1 - Question 4 - HSN forum

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1994 - Paper 1 - Question 4


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#1 Ally

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Posted 20 May 2004 - 12:18 PM

Hi could someone help me with this question:

Differentiate sin³x with respect to x

Hence find ∫ sin²xcosx dx.


I know how to differentiate it, but I can't get the second part

The answer given is 1/3 sin³x

Could someone help...?

#2 Ally

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Posted 20 May 2004 - 12:26 PM

I think it's actually Question 17 and not Question 4

#3 james1

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Posted 20 May 2004 - 12:30 PM

Is this for Higher.

It looks like a C level AH question using integration by substituiton. ohmy.gif

#4 Ally

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Posted 20 May 2004 - 12:36 PM

Yeh its definately higher.

Any tips on how to working it out...?

#5 AndyW

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Posted 20 May 2004 - 12:47 PM

I think this is beyond Higher level:

y = (sin x)^3

let u = sin x

dy/dx = dy/du du/dx
= 3(sin x)^2 * cos x

integral.gif sin²xcosx dx

Note that the term you're integrating is just 1/3 of what you found dy/dx to be previously.

Therefore

integral.gif sin²xcosx dx = 1/3 y = 1/3 (sin^3 x)

#6 Ally

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Posted 20 May 2004 - 12:52 PM

Is it?

It was the last question in the paper so maybe thats why it's so hard.

But still, hopefully no AH questions will come up 2morrow!

#7 Ally

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Posted 20 May 2004 - 01:00 PM

Did this used to be in the Old Higher Course?

If it is, then, maybe thats the disadvantage of doing the really old papers

#8 George

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Posted 20 May 2004 - 01:04 PM

I think it is just a hard Higher question.

I can't see anything specific in the current Arrangements, but you are basically being expected to notice that integrating sin²xcosx dx is related to what you previously worked out.





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