I have the solution to a problem.
$$f(t)=t, \pi ≤t<\pi$$
The solution to coeffisient b_n is: $$ b_n= \frac{1}{\pi}\displaystyle\int_{-\pi}^\pi f(t)sin(nt) dt=\frac{2}{\pi}\displaystyle\int_{0}^\pi f(t)sin(nt) dt$$
so $$\frac{1}{\pi}$$ becomes $$\frac{2}{\pi}$$ and $$\int_{-\pi}^\pi f(t)sin(nt) dt$$ becomes $$\int_{0}^\pi f(t)sin(nt) dt$$
Can someone please explain?
The product of two odd functions, such as $f(t)=t$ and $\sin(nt)$, is an even function,
and the integral of an even function over a symmetric interval, such as $[-\pi,\pi],$
is twice the value of its integral over the interval $[0,\pi]$.