Using the Parseval Identity to compute $ \sum_{n=1}^{+ \infty} \frac{1}{(4n^2-1)^2}$ 
Parseval's Identity: For continuous $f: [- \pi , \pi] \to \mathbb{R}$ 
$$ \sum_{n=- \infty}^{+ \infty} |c_n|^2 = \frac{1}{2 \pi} \int_{ - \pi}^{ \pi} |f(x)|^2dx, \text{ where } c_n = \frac{1}{2 \pi} \int_{- \pi}^{ \pi} f(x)e^{-inx}dx$$

Problem: Using Parseval's Identity, consider the even function $f:= \frac{1}{2} - \frac{\pi}{4} \sin \left( \frac{x}{2} \right), \  x \in [0, \pi] $ to compute the sum: $$\sum_{n =1}^{+ \infty} \frac{1}{(4n^2-1)^2} $$

This is my first attempt to work with Parseval's Identity and Fourier coefficients. I did read this similar question beforehand, but it didn't help me to complete this task  Use Fourier series for computing $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$. 
My approach: I will merely write down the results, I did all the calculations by hand on paper and then double-checked by using Mathematica. Writing out the entire process would only make this post blow up in length further. 
To get the Fourier Coefficients I did integrate using only the definition as above, I first did look at the case for $n=0$. I got $$ c_0 = \frac{1}{2}$$
for the case $ n \neq 0$ I applied integration by parts two times and after a pretty tedious process I did end up with $$c_n = \frac{ in \cos (n \pi)}{4n^2-1}= \frac{in (-1)^n}{4n^2-1} $$
Which was a very satisfying result, because now I have $$|c_n|^2 = \frac{n^2}{(4n^2-1)^2}$$which somewhat looks very close to what I am supposed to get, except for that annoying $n^2$ in the numerator. 
Finally evaluating $$ \frac{1}{2 \pi} \int_{- \pi}^\pi |f(x)|^2dx = \frac{1}{32}(8+ \pi^2)$$
Doubts and Questions: 


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*I have a feeling that my method of integrating was wrong, because I did not make use of $f$ being even which would mean that $f(-x)=f(x)$ and therefore integration from $- \pi, \pi$ would result to be twice the original integral with bounds from 0 to $2 \pi$

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*I did not make use of the statement $f$ is even, because when I plot $f$ it does not look even at all to me, so I wonder about this statement in general. I suppose it has something to do with the definitions of Fourier Coefficients and their application to $A$-periodical Functions. 


*If the above procedure would happen to be right, how could I get the final statement? More precisely formulated, how would I possible get from $$ \sum_{- \infty}^{+ \infty} \frac{n^2}{(4n^2-1)^2}$$ to the Sum that starts from $1$ and ends at $\infty$ ?
 A: Making use of the fact that $f$ is an even function, we extend $f$ as follows:  \begin{align}f:& \begin{cases} [0, \pi] &\longrightarrow \mathbb{R} \\x  & \longmapsto  \displaystyle \frac{1}{2} - \frac{\pi}{4} \sin \left( \frac{x}{2}\right) \end{cases} \tag{given} \\\\\\ f:& \begin{cases} [ - \pi, \pi] & \longrightarrow \mathbb{R} \\ x & \longmapsto \begin{cases} f(x), & 0 \leq x \leq \pi \\ f(-x), & - \pi \leq x \leq 0  \end{cases} \end{cases} \tag{extended} \end{align} 
With this definition of $f$ we have the property that for $ n \neq 0$ (for $n = 0$ analogous) 
\begin{align}c_n = \frac{1}{2 \pi} \int_{- \pi}^{ \pi} f(x) e^{-inx}dx = \frac{1}{2 \pi} \int_{- \pi}^0 \underbrace{f(x)}_{=f(-x)} e^{-inx}dx + \int_0^\pi f(x)e^{inx}dx \end{align}
A simple substitution $u=-x$ for the first integral leads to:
 \begin{align}c_n = \frac{1}{2 \pi} \int_0^\pi f(x) \underbrace{\left( e^{inx}+ e^{-inx} \right)}_{= 2 \cos (nx)}dx \end{align}
which would finally lead to the correct, symmetric Fourier coefficients \begin{align} c_n = \frac{1}{8n^2-2}= \frac{1}{2} \cdot \frac{1}{4n^2-1} \text{ with } c_{-n}=c_n\end{align}
Computing in a likewise manner $$ \frac{1}{ 2 \pi} \int_{- \pi}^\pi |f(x)|^2dx = \frac{\pi ^2-8}{32} $$ leads to the final result: $$ \sum_{n=1}^\infty \frac{1}{4n^2-1}= \frac{\pi^2-8}{16} $$
