Solution:

(i) For the half-range sine series,

Let, $ f(x)=\sum_{n=1}^{\infty} b_{n} \sin n x \\ $

Then, $ b_{n}=\frac{2}{\pi} \int_{0}^{\pi} f(x) \sin n x d x=\frac{2}{\pi}\left[\int_{0}^{\pi / 2} x \sin n x d x+\int_{\pi / 2}^{\pi}(\pi-x) \sin n x d x\right] \\ $

$$ \begin{aligned} &=\frac{2}{\pi}\left[x\left(-\frac{\cos n x}{n}\right)-1 \cdot\left(-\frac{\sin n …