Evaluating a summation of inverse squares over odd indices
Note that $$\sum_{n \text{ is even}} \dfrac1{n^2} = \sum_{k=1}^{\infty} \dfrac1{(2k)^2} = \dfrac14 \sum_{k=1}^{\infty} \dfrac1{k^2} = \dfrac{\zeta(2)}4$$ Also, $$\sum_{k=1}^{\infty} \dfrac1{k^2} = \sum_{k \text{ is odd}} \dfrac1{k^2} + \sum_{k \text{ is even}} \dfrac1{k^2}$$ Hence, $$\sum_{k \text{ is odd}} \dfrac1{k^2} = \dfrac34 \zeta(2)$$
This can be shown in a similar way to Euler's proof of $\zeta(2) = \frac{\pi^2}{6}$, which starts with the function $\frac{\sin(x)}{x}$ (i.e. the sinc function). Here we start with the cosine function which can be expressed as the infinite product
\begin{align} \cos(x) &= \prod_{n=1}^\infty \left(1-\frac{4x^2}{\pi^2(2n-1)^2}\right) \\ &= \left(1- \frac{4x^2}{\pi^2}\right)\left(1- \frac{4x^2}{9\pi^2}\right)\left(1- \frac{4x^2}{25\pi^2}\right) ... \\ &=1-x^2\left(\frac{4}{\pi^2}+\frac{4}{9\pi^2}+\frac{4}{25\pi^2}+...\right)+... \end{align}
$\cos(x)$ can also be expressed by the following Maclaurin series expansion:
$$\cos(x) = \sum_{n=1}^\infty \frac{(-1)^n}{(2n)!}x^{2n} = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!} +...$$
Comparing the $x^2$ coefficients gives:
$$-\frac{1}{2!} = -\frac{4}{\pi^2}\left(1+\frac{1}{9} + \frac{1}{25} + ...\right)$$
Thus,
$$\sum_{n=1}^\infty \frac{1}{(2n-1)^2} = \frac{\pi^2}{8}$$
Here, we present a way forward that does not require prior knowledge of the value of the series $\sum_{n=1}\frac{1}{n^2}=\frac{\pi^2}{6}$, the Riemann-Zeta Function, or dilogarithm function. Rather, we apply straightforward analysis that includes application of the residue theorem.
To that end, note that we can write the series of interest as
$$\begin{align} \sum_{n=0}^\infty \frac{1}{(2n+1)^2}&=\sum_{n=1}^N \int_0^1 x^{2n}\,dx\int_0^1y^{2n}\,dy\\\\ &=\int_0^1\int_0^1 \sum_{n=0}^\infty(x^2y^2)^n\,dx\,dy\\\\ &=\int_0^1\int_0^1 \frac{1}{1-x^2y^2}\,dx\,dy\\\\ &=\frac12\int_0^1\frac{\log(1+x)-\log(1-x)}{x}\,dx\tag 1 \end{align}$$
Then, we enforce the substitution $x\to \frac{x-1}{x+1}$ in $(1)$ to obtain
$$\frac12\int_0^1\frac{\log(1+x)-\log(1-x)}{x}\,dx=\int_1^\infty \frac{\log(x)}{x^2-1}\,dx\tag 2$$
Next, enforcing the substitution $x\to 1/x$ in $(2)$ reveals
$$\int_1^\infty \frac{\log(x)}{x^2-1}\,dx=\int_0^1 \frac{\log(x)}{x^2-1}\,dx \tag 3$$
Adding $(2)$ and $(3)$ and dividing by $(2)$ yields
$$\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac12\int_0^\infty \frac{\log(x)}{x^2-1}\,dx$$
Moving to the complex plane, we evaluate the integral $J$ defined by
$$J=\oint_C \frac{\log^2(z)}{z^2-1}\,dz$$
where $C$ is the classical keyhole contour with (i) the branch cut along the non-negative real axis and (ii) with deformations around $z=1$. Applying the residue theorem, it is easy to see that $J=i\pi^3$. Therefore, we find that $$\begin{align} J&=i\pi^3\\\\ &=\int_{0}^{\infty}\frac{\log^2(x)}{x^2-1}\,dx-\text{PV}\int_0^\infty \frac{\left(\log(x)+i2\pi\right)^2}{x^2-1}\,dx\\\\ &=-i4\pi\int_0^\infty \frac{\log(x)}{x^2-1}\,dx\\\\ &+\color{blue}{(4\pi^2)\text{PV}\left(\int_0^\infty \frac{1}{x^2-1}\,dx\right)}\\\\ &+\color{red}{(4\pi^2)\lim_{\epsilon \to 0^+}\int_{\pi}^{2\pi} \frac{1}{(1+\epsilon e^{i\phi})^2-1}\,(i\epsilon e^{i\phi})\,d\phi}\\\\ &=-i4\pi\int_0^\infty \frac{\log(x)}{x^2-1}\,dx+\color{blue}{0}+\color{red}{i2\pi^3}\tag 4 \end{align}$$
Finally, solving $(4)$ for the integral of interest yields
$$\frac12\int_0^\infty \frac{\log(x)}{x^2-1}\,dx=\frac{\pi^2}{8}$$
and hence we find that the series of interest is
$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac{\pi^2}{8}}$$