# Fourier series How to integrate SinxCosnx?

1. May 5, 2009

### hydralisks

1. The problem statement, all variables and given/known data
Compute the fourier cosine series for given function:

f(x)=sinx 0<x<pi

2. Relevant equations

for cosine series of f(x) on [0,T]... use this general equation:
http://mathworld.wolfram.com/FourierCosineSeries.html

3. The attempt at a solution

so I get:

a0 = (2/pi) * integral(sinxdx) with bounds 0 to pi = 4/pi

but then.. when I try to compute an

I get
an=(2/pi) * integral(sinx*cosnx*dx) with bounds 0 to pi
How do I integrate sinxcosnx?

Last edited: May 5, 2009
2. May 5, 2009

### Cyosis

How did you go from f(x)=e^x to f(x)=sin(x)? Either way use the product-to-sum identity to write sin(x)cos(nx) as a sum of sine functions or express the sine and cosine functions in terms of complex exponentials.

3. May 5, 2009

### hydralisks

Ah, sorry I'm becoming delusional from doing too much work in one day. It's fixed now.

I do get sin(x)cos(nx) right? Or am I doing something wrong

4. May 5, 2009

### Cyosis

Yes you do get sin(x)cos(nx) with the edited f(x). Now use $$\sin \theta \cos \varphi = {\sin(\theta + \varphi) + \sin(\theta - \varphi) \over 2}$$

5. May 5, 2009

### hydralisks

Ah, thanks so much. Sorry I actually have another question though,

I am asked to compute the Fourier series for the following 2 part function:

f(x)=1 -2<x<0
f(x)=x 0<x<2

I'm supposed to do this using the "Euler formulas" not the cos/sin formulas.

However, I'm not sure how this two part thing works. When trying to find an, do I just do the integral of part 1 + integral of part 2?

so...
an = [(1/2)*integral(1*cos(n*pi*x/2)) from -2 to 0] + [(1/2)*integral(x*cos(n*pi*x/2)) from 0 to 2]

6. May 5, 2009

### Cyosis

You're splitting the integral up correctly, but you said you're supposed to use the exponential form of the Fourier-transform after which you use the cosine form instead (confusing). Do it again using the exponential form.