Showing that the graph of $\frac{\cos(\pi x/2)}{1-x^2}$ decreases over the interval $(0,2)$. How would one justify the behaviour of the graph of 
$$\frac{\cos(\frac{\pi}{2}x)}{1-x^2}$$ 
over the interval $(0,2)$?
More precisely, how can it be shown that the graph is decreasing over that interval?

The derivative of $f(x)$ takes the following form
\begin{equation}
 f'(x) = \frac{g(x)}{(1-x^2)^2}
\end{equation}
\begin{equation}
 g(x) = -\frac{\pi}{2} \sin (\frac{\pi}{2}x)(1-x^2)+2x( \cos (\frac{\pi}{2}x))
\end{equation}
PS: The function $f(x)$ is not defined at $1$
Since the denominator of $f'(x) > 0$ on $[0,1[ \cup ]1,2]$, we will study $g(x)$. Get it's derivative
\begin{equation}
 g'(x) = -\frac{\pi^2}{4} \cos (\frac{\pi}{2}x)(1-x^2)
 +
 \pi x \sin (\frac{\pi}{2}x)
 +
 2\cos (\frac{\pi}{2}x)
 -\pi x \sin (\frac{\pi}{2}x)
\end{equation}
which is 
\begin{equation}
 g'(x) = -\frac{\pi^2}{4} \cos (\frac{\pi}{2}x)(1-x^2)
 +
 2\cos (\frac{\pi}{2}x)
\end{equation}
Factorize to get
\begin{equation}
 g'(x) = [-\frac{\pi^2}{4}(1-x^2)
 +
 2]\cos (\frac{\pi}{2}x)
\end{equation}
You could further factorize as 
\begin{equation}
 g'(x) = \big(x - \sqrt{1 - \frac{8}{\pi^2}} \big)\big(x + \sqrt{1 - \frac{8}{\pi^2}} \big)\cos (\frac{\pi}{2}x)
\end{equation}

On the interval $[0,1]$
In the interval $[0,1]$, we know that both $\cos(\frac{\pi}{2})$ and $x + \sqrt{1 - \frac{8}{\pi^2}}$ are positive. So in $[0,1]$ the sign of $g'(x)$ depends on $x - \sqrt{1 - \frac{8}{\pi^2}}$. Obviously we get that $g'(x) < 0$ in $[0,x_0]$ and $g'(x) > 0$ in $[x_0,1]$ where $x_0 = \sqrt{1 - \frac{8}{\pi^2}}$. So there is a minimum at $x_0$. But $g(0) = g(1) = 0$. So $g(x) < 0$ in $[0,1]$. 

On the interval $[1,2]$
In the interval $[1,2]$, we have that both $x + \sqrt{1 - \frac{8}{\pi^2}}$  and $x + \sqrt{1 + \frac{8}{\pi^2}}$ are positive so the sign of $g'(x)$ depends on $\cos \frac{\pi}{2}x$ which is negative in $[1,2]$. Hence, $g(x)$ is decreasing decreasing in $[1,2]$, but $g(1) = 0$, hence $g(x)$ is also negative in $[1,2]$. 

Concluding on $[0,2]$
We conclude that in $[0,2]$, 
\begin{equation}
 g(x) \leq 0
\end{equation}
Hence $f(x)$ is decreasing in $[0,2]$, excluding $x = 1$.
