I want to prove that for the function
$$ f(x) = \begin{cases} \frac{π}{4} & 0 <x < π \\ \frac{-π}{4} & -π < x < 0 \end{cases} $$
the Fourier series is: $$f(x) = \sin x + \frac{\sin3x}{3} + \frac{\sin5x}{5} + \frac{\sin7x}{7}+\dots $$
[My Attempt] $$f(x) = a_0\sum_{i=0}^∞ a_n\cos(nx) + b_n\sin(nx) $$ $$a_0 = \frac{1}{π}\int_{-\pi}^{+\pi}f(x)\cos(nx) dx$$ $$b_n = \frac{1}{π}\int_{-\pi}^{+\pi}f(x)\sin(nx) dx$$ but I don't know how to prove from here.
What should be the next step?
Note that $f(x)$ is odd, hence all $a_n$ terms vanish. Thus, the Fourier series admits the form: $$f(x)=\sum_{n=1}^{\infty} b_n\sin(nx) \tag{1}$$ Since $f(x)$ is odd, it follows that $f(x)\sin(nx)$ is even, since the product of two odd functions is even. Hence, it follows that: $$\begin{align}b_n&=\frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin(nx)~dx\\&=\frac{2}{\pi}\int_0^{\pi} f(x)\sin(nx)~dx\\&=\frac{2}{\pi}\int_0^{\pi} \frac{\pi}{4}\cdot \sin(nx)~dx\\&=\frac{1}{2}\int_0^{\pi}\sin(nx)~dx \end{align}$$ I will leave the rest as an exercise. Once you have computed $b_n$ for all $n\in \mathbb{N}$, substitute your result into $(1)$, then you are done (Assuming you have computed the integral correctly).