# Theoretical Integration

1. Feb 7, 2006

### jamesbob

I'm stuck on explaining this. Does anybody know how to help?

(a) By writing $$\cos^nx = cos^{n-1}xcosx$$ use integration by parts to show that

$$\int \cos^nxdx = \cos^{n-1}xsinx + (n-1) \int \sin^2xcos^{n-2}xdx.$$

(b) Using the result of part (a) derive the reduction formula

$$n\int \cos^nxdx = \cos^{n-1}x\sinx + (n-1) \int \cos^{n-2}xdx.$$​

My Working:

(a) All i got so far is

u = cosx dv/dx =cos^{n-1}x

du/dx = -\sinx v = \int \cos^{n-1}x

Last edited: Feb 7, 2006
2. Feb 7, 2006

### StatusX

You usually don't want to pick a function you can't immediately integrate as dv/dx, because then you have a nested integral, which is no fun. Try switching your choices of u and dv/dx.

3. Feb 7, 2006

### jamesbob

ok i swithced u and v and got:

$$u = \cos^{n-1} \mbox{ and so } \frac{du}{dx} = \frac{-sin(n-1)x}{(n-1)^2}$$

is this right so far?

4. Feb 7, 2006

### StatusX

I don't think so, although I don't really understand what you've written for du/dx. Just use the chain rule to differentiate (cos(x))^(n-1).

5. Feb 7, 2006

### jamesbob

ok i got the first bit. using the chain rule works, thanks. but how do you do (b)? how can multiplying by n cancel out the sin^2x in the integration?

6. Feb 7, 2006

### StatusX

Use sin2(x)=1-cos2(x). Also, I think there's a mistake in the formula you posted. There should be a sin(x) in the first term on the right side.

Last edited: Feb 7, 2006
7. Feb 7, 2006

### jamesbob

hmm, i can't see why using sin^2x = cos^2x - 1 helps, sorry. Could you explain it further please?