**Differentiate sin³x with respect to x**

Hence find ∫ sin²xcosx dx.

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...?

Started by Ally, May 20 2004 12:18 PM

7 replies to this topic

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...?

Hence find ∫ sin²xcosx dx.

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

The answer given is

Could someone help...?

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Posted 20 May 2004 - 12:26 PM

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

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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.

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

Posted 20 May 2004 - 12:36 PM

Yeh its definately higher.

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

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

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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

sin²xcosx dx

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

Therefore

sin²xcosx dx = 1/3 y = 1/3 (sin^3 x)

y = (sin x)^3

let u = sin x

dy/dx = dy/du du/dx

= 3(sin x)^2 * cos x

sin²xcosx dx

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

Therefore

sin²xcosx dx = 1/3 y = 1/3 (sin^3 x)

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!

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!

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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

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

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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.

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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