b_n=\frac 22\int_0^2f(x)\,\sin(n\pi x/2)\,dx=\int_0^12\sin(n\pi x/2)\,dx+\int_1^2x\,\sin(n\pi x/2)\,dx=\frac 4{n\pi}(1-\cos(n\pi/2))+\frac 2{n^2\pi^2}(n\pi\cos(n\pi/2-2\sin(n\pi/2)-2\pi(\cos(n\pi))))