prove that $\int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx < \frac{\pi^2}{64}$ prove that $\int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx <  \frac{\pi^2}{64}$
I showed that in 
$$ \forall x \in [0,\frac{\pi}{4}] \quad  \int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx \le \int_0^\frac{\pi}{4} \frac{1-\cos(\frac{\pi}{4})}{\frac{\pi}{4}} \,dx = 1 -\frac{\sqrt2}{2} $$
but it's not good enough

Another way forward is to recall the Taylor series for the cosine
$$\cos x =\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \tag 1$$
Noting that for $0\le x\le \pi/4$ the terms of the series in $(1)$ alternate signs and are decreasing monotonically. Therefore, we have
$$\cos x \ge 1-\frac12 x^2$$
so that
$$\frac{1-\cos x}{x}\le \frac12 x$$
Finally, we see that 
$$\begin{align}
\int_0^{\pi/4}\frac{1-\cos x}{x}\,dx &\le \int_0^{\pi/4} \frac12 x\, dx\\\\
&=\frac{\pi^2}{64}
\end{align}$$
as was to be shown!
