# Homework Help: Integration proof

1. Apr 15, 2013

### synkk

I'm almost out of high school and have been trying to do some "harder" proofs, anyone, I'm not quite sure on how to proceed on this:

prove that for all k:

$$\displaystyle \int_0^{2\pi} (cos^{2k}x - sin^{2k} x) \ dx = 0$$

If anyone could start me off as I'll I have in my head is "try to express it into something you can integrate", but having no luck.

Last edited: Apr 15, 2013
2. Apr 15, 2013

### SteamKing

Staff Emeritus
It's not clear what k is limited to. Is k all integers, all reals, etc? If k is an integer, will its range include negative values? What about k = 0?

3. Apr 15, 2013

### synkk

The parts before this question asked me to integrate cos^6(x) and sin^6x using identities from cos^6(x) - sin^6(x) and cos^6x + sin^6x which I done successfully. At the end of the question it says: "You might like to consider how you would prove that $$\displaystyle \int_0^{2\pi} (cos^{2k}x - sin^{2k} x) \ dx = 0$$ for all k, without having to derive a new identity for each value of k.

4. Apr 15, 2013

### tiny-tim

hint: use symmetry

5. Apr 17, 2013

### synkk

Could you expand on that? I tried rewriting cos in terms of sin, but still pretty stuck

thanks

6. Apr 17, 2013

### tiny-tim

cos2k(π/2 - x) - sin2k(π/2 - x) = … ?

7. Apr 17, 2013

### synkk

= $sin^{2k}x - cos^{2k}x = -(cos^{2k}x - sin^{2k} x)$

I feel so dumb for not getting this :(

I tried for around 15 minutes playing around with what you gave me to show it's 0 but to no avail... I'll try again tomorrow and post if I've made progress. Thanks for your patience.

8. Apr 17, 2013

### Curious3141

EDIT: Ignore my previous post, I'd made an error.

Last edited: Apr 18, 2013
9. Apr 17, 2013

### Ray Vickson

Look at the graphs of sin(x) and cos(x). Can you get one of the graphs by shifting the other one to the right or the left?

10. Apr 19, 2013

### synkk

π
yes cos(x) = sin(x + π/2)
also sin(x) = cos(x - π/2)

I can see why the integral is 0, by drawing both of the graphs and shading in the required area, but I'm having real trouble formalising a proof to actually put onto paper :\

11. Apr 19, 2013

### tiny-tim

try substituting y = π - x, or y = π/2 - x (and dy = -dx), and seeing what it does to the integral of one of your shaded parts