Is there any symmetry I can use to find this Fourier sine series?

[richyw]

Homework Statement

I am going over a practice exam, and I need to find the FSS of f(x)=x(\pi^2-x^2)

Homework Equations

f(x) \sim \sum^\infty_{n=1}a_n sin\left(\frac{n \pi x}{L}\right)
a_n=\frac{2}{L}\int^L_0 f(x)sin\left(\frac{n\pi x}{L}\right)dx

The Attempt at a Solution

I think I need to integrate\frac{2}{\pi}\int^\pi_0 x(\pi^2-x^2)sin(n\pi)dxwhich is two integrals, the first one would need me to use IBP once, and the second one would need me to use IBP three times. This is on a practice exam (and my exam is in an hour), so I am guessing that this integral is easier if I can find some symmetry in it. Is this true?

 

[asdf12312]

You can indeed use symmetry. Graph f(x) on your calc: what do you notice? remember if x(t)=x(-t) then there is even symmetry, x(t)=-x(-t) means it has odd symmetry.

 

[LCKurtz]

Homework Statement

I am going over a practice exam, and I need to find the FSS of f(x)=x(\pi^2-x^2)

Homework Equations

f(x) \sim \sum^\infty_{n=1}a_n sin\left(\frac{n \pi x}{L}\right)
a_n=\frac{2}{L}\int^L_0 f(x)sin\left(\frac{n\pi x}{L}\right)dx

The Attempt at a Solution

I think I need to integrate\frac{2}{\pi}\int^\pi_0 x(\pi^2-x^2)sin(n\pi)dxwhich is two integrals, the first one would need me to use IBP once, and the second one would need me to use IBP three times. This is on a practice exam (and my exam is in an hour), so I am guessing that this integral is easier if I can find some symmetry in it. Is this true?

The fact that the function is odd is the reason you can use the half range sine expansion in the first place. That is why there are no cosine terms. You just need to bite the bullet and do the integration by parts. Don't do two separate integrals though. Write your integrand as$$
(\pi^2x-x^3)sin(n\color{red}x)$$and do it all at once. Note your typo correction.