Fourier series How to integrate SinxCosnx?

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Homework Statement

Compute the Fourier cosine series for given function:

f(x)=sinx 0<x<pi

Homework Equations

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

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 a_n

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

 

[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.

 

[hydralisks]

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.

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

 

[Cyosis]

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

 

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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 a_n, do I just do the integral of part 1 + integral of part 2?

so...
a_n = [(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]

 

[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.