What will be the value of $\displaystyle\int_{0}^{1} \frac{1-\cos(x)}{x} dx$ for the first three decimal digit? $$\displaystyle\int_{0}^{1} \frac{1-\cos(x)}{x} dx$$
Can I somehow write the Taylor-series of $\frac{1-\cos(x)}{x}$?

Observe we have
\begin{align}
1-\cos x = \sum^\infty_{n=1} (-1)^{n+1}\frac{x^{2n}}{(2n)!}
\end{align}
which means
\begin{align}
\int^1_0 \frac{1-\cos x}{x} \ dx = \sum^\infty_{n=1}(-1)^{n+1}\frac{1}{(2n)!}\int^1_0 x^{2n-1}\ dx = \sum^\infty_{n=1}(-1)^{n+1}\frac{1}{(2n)!}\frac{1}{2n}.
\end{align}
Hence
\begin{align}
\left|\sum^\infty_{n=N}(-1)^{n+1}\frac{1}{(2n)!2n}\right| \leq \sum^\infty_{n=N} \frac{1}{(2n)! 2n} \leq \sum^\infty_{n=N}\frac{1}{2^{2(n+1)}} = \sum^\infty_{n=N}\frac{1}{4^{n+1}} = \frac{3}{4^{N}} 
\end{align}
when $N\geq 4$. In particular, if you want three digit accuracy, all you have to do is choose $N$ such that 
\begin{align}
\frac{3}{4^N}<10^{-3}
\end{align}
which means you need $N\geq 6$. Hence sum up six terms to get at least 3 decimal places of accuracy. 
