# Derivative of cos^(n-1) x

1. Oct 17, 2009

### Patjamet

I'm working on a Maths assignment and one of the problem solving questions is to prove:
$$\int cos^{8}(x)dx=\frac{1}{8}cos^{7}(x).sin(x)+\frac{7}{48}cos^{5}(x).sin(x)+\frac{35}{192}cos^{3}(x).sin(x)+\frac{35}{128}cos(x).sin(x)+\frac{35}{128}(x)+c$$

I have looked through my text book and found a good example which will help me, the book uses a thing they are calling a recursive formula. I've done some research and I've seen the same formula from lots of different information sources.

Exhibit A: http://calc101.com/deriving_reduction_2.html

[STRIKE]They have all given demonstrations on how to get the formula, but when I try to do it myself I run into a problem when differentiating the first function $$cos^{n-1}(x)$$[/STRIKE]

Nevermind I was integrating haha, but still; obviously I am rusty on my differentiation techniques so if someone could check this for me that would be fantastic (mainly concerned with my procedure for communication marks etc.) :)

$$y=cos^{n-1}(x)$$

Allow cos(x) to equal u

$$\frac{dy}{dx}=u^{n-1}.............1$$

$$\frac{du}{dx}=-sin(x)$$

$$\frac{du}{-sin(x)}=dx..............2$$

Sub 2 into 1

$$\frac{dy}{\frac{du}{-sin(x)}}=u^{n-1}$$

$$\frac{dy}{du}=(n-1)u^{n-1-1}.-sin(x)$$

$$\frac{dy}{du}=(n-1)u^{n-2}.-sin(x)$$

But u = cosx

$$\frac{dy}{du}=(n-1)cos^{n-2}(x).-sin(x)$$

2. Oct 17, 2009

### HallsofIvy

Staff Emeritus
No, you haven't done the derivative yet. $y= u^{n-1}$
and then $dy/dx= (n-1)u^{n-2} [/quote]$$\frac{du}{dx}=-sin(x)$$ Why are you doing this? Use the chain rule: (du/dx)= (du/dy)(dy/dx) That gives you [itex])n-1)u^{n-2}(-sin(x))= -(n-2)sin(x)cos^{n-2}(x) But I see no reason to ask about the "n-1" power. For your problem you want to look at [itex]cos^n(x)$. Again, letting u= cos(x), $y= u^n$ so $dy/du= n u^{n-1}$ and du/dx= - sin(x). So $dy/dx= (dy/du)(du/dx)= n u^{n-1}(-sin(x))= -n sin(x)cos^{n-1}(x)$.

With a little practice, you should be able to do that with actually writing down the "u" substitution: to differentiate $cos^n(x)$ think "The "outer function" is a power so the derivative is $n cos^{n-1}(x)$ and then I multiply by its derivative, -sin(x)".

3. Oct 17, 2009

### Patjamet

Doh, the chain rule.

Thank you very much, looks like I have alot of revision to do!