Integrate cos(2x)^N

[lil_luc]

Hi there,

I am having a hard time figuring out how to integrate cos(2x)^N where N=0,1,2....

Can anyone give me and hints/tips on how to go about integrating this?
Thanks!

[n!kofeyn]

I haven't rechecked my work yet, but I think this works. Also, I'm assuming you meant \int \cos^n(2x)\,dx. Rewrite the integral as
\int \cos^{n-1}(2x)\cos(2x)\,dx
Then use integration by parts. Once you do this, use integration by parts again to evaluate the integral that's left over. You will get constant*I = stuff - (another constant)*I, where I is the integral left over after the first integration by parts. Now solve for I.

I hope this works. I'll do a double check later. Let me know how this works out.

[lil_luc]

Thank you for the hint. I will try this out. I didn't think to seperate it like that and do integration by parts. But I will give it a shot and let you know how that goes for me! =S

[HallsofIvy]

First make the substitution u= 2x and the problem becomes just
\frac{1}{2}\int cos(u)du
so the "2" inside the cosine is irrelevant.

If n is odd, say, n= 2k+1, it is easy. Factor one cosine out for the 'du' and make the substitution y= cos(u).

If n is even, say, n= 2k, that is harder. Use the trig substitution cos^2(u)= (1/2)(1+ cos(2u)) to reduce the power:
\int cos^{2k}(u)du= \frac{1}{2}\int du+ \frac{1}{2}\int cos^k(u)du
= \frac{u}{2}+ \frac{1}{2}\int cos^k(u)du
So you have a recursion relation. I suspect doing the n= 2, 4, 6 (so k= 1, 2, 3) will give you an idea of the pattern and then the recursion relation will allow you to prove that pattern by induction.