Prove some formulas for given infinite sums

Using the formula

$\sum_{n=1}^{\infty} \frac{\cos (nx)}{n^2} = \frac{x^2}{4} - \frac{\pi x}{2} + \frac{\pi^2}{6} \qquad 0 \leq x \leq 2 \pi$

prove that:

$\displaystyle{\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}}$ .
$\displaystyle{\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n-1)^3} = \frac{\pi^3}{32}}$ .

Proof. Taking $x = 0$ in the given formula we have
$\begin{align*} && \sum_{n=1}^{\infty} \frac{\cos (nx)}{n^2} &= \frac{x^2}{4} - \frac{\pi x}{2} + \frac{\pi^2}{6} \\[9pt] \implies && \sum_{n=1}^{\infty} \frac{\cos (0)}{n^2} &= \frac{\pi^2}{6} \\[9pt] \implies && \sum_{n=1}^{\infty} \frac{1}{n^2} &= \frac{\pi^2}{6}. \qquad \blacksquare \end{align*}$
Proof. Starting with the given equation we have for $0 \leq x \leq 2 \pi$ ,
$\begin{align*} && \sum_{n=1}^{\infty} \frac{\cos (nx)}{n^2} &= \frac{x^2}{4} - \frac{\pi x}{2} + \frac{\pi^2}{6} \\[9pt] \implies && \int \sum_{n=1}^{\infty} \frac{\cos (nx)}{n^2} \, dx &= \int \left( \frac{x^2}{4} - \frac{\pi x}{4} + \frac{\pi^2}{6} \right) \, dx \\[9pt] \implies && \sum_{n=1}^{\infty} \left(\int \frac{ \cos (nx)}{n^2} \, dx \right)&= \frac{x^3}{12} - \frac{\pi x^2}{4} + \frac{\pi^2 x}{6} \\[9pt] \implies && \sum_{n=1}^{\infty} \frac{\sin (nx)}{n^3} &= \frac{x^3}{12} - \frac{\pi x^2}{4} + \frac{\pi^2 x}{6}. \end{align*}$

Take $x = \frac{\pi}{2}$ , then we have

$\begin{align*} && \sum_{n=1}^{\infty} \frac{\sin \left( \frac{\pi}{2} \cdot n \right)}{n^3} &= \frac{ \left(\frac{\pi}{2} \right)^3}{12} - \frac{ \pi \left( \frac{\pi}{2} \right)^2}{4} + \frac{\pi^2 \left( \frac{\pi}{2} \right)}{6} \\[9pt] \implies && \sum_{n=1}^{\infty} \frac{\sin \left( n \pi - \frac{\pi}{2} \right)}{(2n-1)^3} &= \frac{\pi^3}{32} &(\sin (n \pi) = 0 \text{ for even }n)\\[9pt] \implies && \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n-1)^3} &= \frac{\pi^3}{32}. \qquad \blacksquare \end{align*}$