# Integral and Fourier serie

1. Aug 22, 2013

### Mathitalian

1. The problem statement, all variables and given/known data

Let $$u(t)=2-\cos(t)+\sin(2t)- \cos(3t)+ \sin(4t)$$

Evaluate:

$$\int_0^{2\pi}u^2(t)\mbox{d}t$$

2. Relevant equations

3. The attempt at a solution

Sorry, I don't have any idea :(... As I can see

$$\int_0^{2\pi}u^2(t)\mbox{d}t$$

is similar to the first term of Fourier Serie of the function u^2(t)...

2. Aug 23, 2013

### MisterX

You might consider using Parseval's theorem.

3. Aug 23, 2013

### HallsofIvy

Staff Emeritus
I don't see any reason to think about "Fourier Series" at all- just integrate directly:
$\int_0^{2\pi} 4+ cos^2(t)+ sin^2(2t)+ cos^2(3t)+ sin^2(4t) dt$
is easy.
$2\int_0^{2\pi}-2cos(t)+ 2sin(2t)- 2cos(3t)+ 2sin(4t)dt$
$2\int_0^{2\pi}- cos(t)sin(2t)- cos(t)cos(3t)+ cos(t)sin(4t)dt$
$2\int_0^{2\pi}-sin(2t)cos(3t)+ sin(2t)sin(4t)- cos(3t)sin(4t)dt$

You may need the trig identities for cos(a)cos(b), sin(a)sin(b), and sin(a)cos(b).

4. Aug 23, 2013

### stevenb

HallsOfIvy,

I think you just squared each term, rather than squaring the whole expression?

5. Aug 23, 2013

### LCKurtz

..along with the fact that $u(t)$ is its own Fourier Series.

6. Aug 23, 2013

### Mathitalian

So

$$\int_{0}^{2\pi}u^2(t)\mbox{d}t= 12\pi$$

right? :)

7. Aug 23, 2013

### Dick

Right. But if you think about orthogonality relations, you only have to evaluate the first integral Halls gave in post 3. The rest are obviously 0. And that one doesn't take much calculation either if you remember that the 'average value' of sin(nt)^2 or cos(nt)^2 where n>1 is 1/2.

Last edited: Aug 23, 2013
8. Aug 23, 2013

### LCKurtz

Actually, you don't have to evaluate any integrals. The Fourier coefficients are sitting right there in the function u(t).

9. Aug 24, 2013

### Dick

Sure, if you use Parseval's. I wasn't sure which track the OP was following.