Find the Fourier series of this function, only by using sine functions.
This is not a homework, I'm just practicing different problems for an exam.
I know that all coefficients, except b0 should be 0.
I'm just not sure how to integrate this, because this intervals are confusing me.
Also, I can't find any similar problem on the internet, and there is no similar example solved in the book. Any help would be appreciated.
$f(x) = \begin{cases} x, & x∈[0,\pi] \\[2ex] -x, & x∈[\pi,2\pi] \end{cases}$
Since you want the sine series (if I understand you correctly), you first extend your function to an odd function: $$ \tilde f(x)= \begin{cases} -x & -2\pi<x<-\pi\\ x & -\pi\leq x\leq \pi\\ -x & \pi<x\leq 2\pi \end{cases} $$ Then calculate the Fourier coefficients of this $4\pi$-periodic function as usual: $$ a_n=\frac{2}{4\pi}\int_{-2\pi}^{2\pi}\tilde f(x)\cos\frac{2\pi n x}{4\pi}\,dx=0 $$ since the integrand is odd, and (here we use in the second step that the integrand is even) $$ \begin{aligned} b_n&=\frac{2}{4\pi}\int_{-2\pi}^{2\pi}\tilde f(x)\sin\frac{2\pi n x}{4\pi}\,dx\\ &=\frac{1}{\pi}\int_{0}^{2\pi} f(x)\sin\frac{n x}{2}\,dx=\cdots \end{aligned} $$ insert the definition of $f$ and calculate the integral.