Answer
$\mathop \smallint \limits_0^x \dfrac{{1 - \cos t}}{t}{\rm{d}}t = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 2}}}}{{\left( {2n + 2} \right)\left( {2n + 2} \right)!}}$
Work Step by Step
Write $\dfrac{{1 - \cos t}}{t} = \dfrac{1}{t}\left( {1 - \cos t} \right)$.
From Table 2:
$\cos t = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{t^{2n}}}}{{\left( {2n} \right)!}} = 1 - \dfrac{{{t^2}}}{{2!}} + \dfrac{{{t^4}}}{{4!}} - \dfrac{{{t^6}}}{{6!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $t$.
Thus,
$\dfrac{{1 - \cos t}}{t} = \dfrac{1}{t}\left( {1 - \cos t} \right) = \dfrac{1}{t}\left( {1 - \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{t^{2n}}}}{{\left( {2n} \right)!}}} \right)$, ${\ \ \ }$ converges for all $t$.
$\dfrac{{1 - \cos t}}{t} = \dfrac{1}{t}\left( {\dfrac{{{t^2}}}{{2!}} - \dfrac{{{t^4}}}{{4!}} + \dfrac{{{t^6}}}{{6!}} - \cdot\cdot\cdot} \right)$
$\dfrac{{1 - \cos t}}{t} = \dfrac{t}{{2!}} - \dfrac{{{t^3}}}{{4!}} + \dfrac{{{t^5}}}{{6!}} - \cdot\cdot\cdot = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{t^{2n + 1}}}}{{\left( {2n + 2} \right)!}}$
Taking the definite integral on the series gives
$\mathop \smallint \limits_0^x \dfrac{{1 - \cos t}}{t}{\rm{d}}t = \mathop \sum \limits_{n = 0}^\infty \mathop \smallint \limits_0^x \dfrac{{{{\left( { - 1} \right)}^n}{t^{2n + 1}}}}{{\left( {2n + 2} \right)!}}{\rm{d}}t = \mathop \sum \limits_{n = 0}^\infty \left[ {\dfrac{{{{\left( { - 1} \right)}^n}{t^{2n + 2}}}}{{\left( {2n + 2} \right)\left( {2n + 2} \right)!}}} \right]_0^x$
So, $\mathop \smallint \limits_0^x \dfrac{{1 - \cos t}}{t}{\rm{d}}t = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n + 2}}}}{{\left( {2n + 2} \right)\left( {2n + 2} \right)!}}$.
