# Edexcel C4 Chapter 6 Integration L 34

Discussion in 'Mathematics' started by pipipi, May 17, 2011.

1. ### pipipi

I having a brain melt on this question.

For the second bit the solution shows sin^2 t cos t, to be integrated.
And the next line has the answer to this as sin^3 t .
It has gone a bit too fast for me to keep up there. Can anyone fill in the steps please?

2. ### KYPNew commenter

Hmm ... I think it should be (sin^3 t)/3. It's the inverse chain rule. cos t is the derivative of sin t, so (sin t)^3 differentiates to 3(sin t)^2 * cos t . Hard to explain in writing!

3. ### pipipi

Thanks KYP. The solution does have 1/3 as the coefficient but like you, I'm having trouble typing it correctly.
Will take a look at the inverse chain rule. Thanks!

4. ### briceanusNew commenter

If still in doubt, use the substitution u = sin(t). Not forgetting to find the necessary replacement for dt in terms of du.
Bric

5. ### Betamale

sin^3x is written as often promoted as sinx(1-cos^2x) in C4 and sin^2x as 1/2(1-cos2x)

6. ### beedgeNew commenter

I think some of the responses to your question are a bit more techinical than they need to be. Particularly if you start integrating a product.
If you differentiate something like (sin x)^5 you get 5 (sinx)^4 cos x
So if you have to integrate (sin x)^2 cos x, then you need to recognise that this has come from (sin x)^3 differentiated - more or less.
Derrivative of (sin x)^3 = 3(sinx)^2 cos x
Therefore integral of (sinx)^2 cos x = 1/3 (sin x)^3
I think the textbooks do give a general form of integrating A(sinx)^n cos x - but it seems like one more formula to learn. I wouldn't bother with it.