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
 A: Using the fact that $\sin(x)\le x$ for $x\ge0$,
$$
\begin{align}
\int_0^{\pi/4}\frac{1-\cos(x)}x\,\mathrm{d}x
&=\int_0^{\pi/4}\frac{2\sin^2(x/2)}x\,\mathrm{d}x\\
&=\int_0^{\pi/4}\frac{\sin(x/2)}{x/2}\cdot\sin(x/2)\,\mathrm{d}x\\
&\le\int_0^{\pi/4}1\cdot\frac x2\,\mathrm{d}x\\[4pt]
&=\frac{\pi^2}{64}
\end{align}
$$
A: 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!
A: Hint: $1-\cos(x)=2\sin^2\left(\frac{x}{2}\right)$ and $0 \leq \frac{\sin(t)}{t} < 1$ for all $t \in \left(0,\frac{\pi}{2}\right]$.  If done properly, you should get $\int\limits_0^{\pi/4}\,\frac{1-\cos(x)}{x}\,\text{d}x<\frac{\pi^2}{64}$ (which I think you meant to do, since $\frac{\pi^2}{4}$ is a very weak bound).
A: Since:
$$\frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2\pi^2}\right) $$
over the interval $\left[0,\frac{\pi}{4}\right]$ we have:
$$ \frac{1-\cos x}{x}=\left(\frac{\sin(x/2)}{x/2}\right)^2\cdot\frac{x}{2}\leq\left(1-\frac{x^2}{4\pi^2}\right)^2\frac{x}{2} $$
so:
$$ \int_{0}^{\frac{\pi}{4}}\frac{1-\cos x}{x}\,dx \color{red}{\leq} \int_{0}^{\pi/4}\left(1-\frac{x^2}{4\pi^2}\right)^2\frac{x}{2}\,dx = \frac{\pi^2}{64}-\frac{191\,\pi^2}{3\cdot 2^{18}}.$$
