7 replies to this topic

[Ally]

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...?

[Ally]

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

[james1]

Is this for Higher.

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

[Ally]

Yeh its definately higher.

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

[AndyW]

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)

[Ally]

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!

[Ally]

Did this used to be in the Old Higher Course?

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

[George]

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.