<s>[INST] How to estimate the following integral: $\int_0^1 \frac{1-\cos x}{x}\,dx$ 
How to estimate the following integral?
  $$
\int_0^1 \frac{1-\cos x}{x}\,dx
$$

 [/INST] Since $\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$,
$1-\cos(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{2n}}{(2n)!}$,
so 
$\frac{1-\cos(x)}{x} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{2n-1}}{(2n)!}$.
Therefore
$\begin{align}
\int_0^1 \frac{1-\cos(x)}{x} dx
&= \int_0^1 dx \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{2n-1}}{(2n)!}\\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{(2n)!} \int_0^1 x^{2n-1} dx\\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{(2n)!} \frac1{2n} \\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{2n(2n)!}\\
\end{align}
$
This is an alternating series with terms decreasing in absolute value,
so its sum is between any two consecutive terms.
The first two terms are
$\frac1{4}$ and $\frac{-1}{4\cdot 4!} = \frac{-1}{96}$,
so the result is slightly less than $\frac1{4}$.
</s>