Calculating an easy iterated integral I'm trying to solve the following integral:

$$ \int_{0}^{t}dt_{1}^{'} \int_{0}^{t_1}dt_{2}^{'} \; \sin \; [k(t_1^{'} - t_2^{'})]. $$

The correct answer is:

$$  \frac{\sin(kt) - kt}{k^2}.      $$

My reasoning, however, gives me the following result:
The first inetgral gives me the result:

$$  \int_{0}^{t_1} dt_{2}^{'} \; \sin \; [k(t_1^{'} - t_2^{'})] = \frac{- \cos \; [k(t_1^{'} - t_2^{'})] }{-k} \mid_{0}^{t_1} = \frac{\cos \; [k(t_1^{'} - t_1)]}{k} - \frac{\cos \; (kt_1^{'})}{k}.                     $$

The second integral gives the result:

$$  \int_{0}^{t} dt_{1}^{'} \; \frac{\cos \; [k(t_1^{'} - t_1)]}{k} - \frac{\cos \; (kt_1^{'})}{k} = \frac{\sin \; [k(t - t_1)]}{k^2}  - \frac{\sin \; [k(- t_1)]}{k^2}  - \frac{\sin(kt)}{k^2}.              $$

My final answer is:

$$ \frac{\sin \; [k(t - t_1)]}{k^2}  + \frac{\sin \; [kt_1]}{k^2}  - \frac{\sin(kt)}{k^2}.  $$

But this clearly doesn't match with the correct answer, as given in the resource from where I got the integral.
Any suggestions?
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
&\color{#f00}{%
\int_{0}^{t}\dd t_{1}\int_{0}^{t_{1}}\dd t_{2}
\sin\pars{k\bracks{t_{1} - t_{2}}}} =
\Im\int_{0}^{t}\exp\pars{\ic k t_{1}}\int_{0}^{t_{1}}
\exp\pars{-\ic kt_{2}}\,\dd t_{2}\,\dd t_{1}
\\[3mm] = &\
\Im\int_{0}^{t}\exp\pars{\ic kt_{1}}\,
{\exp\pars{-\ic kt_{1}} - 1 \over -\ic k}\,\dd t_{1} =
{1 \over k}\,\Re\int_{0}^{t}\bracks{1 - \exp\pars{\ic kt_{1}}}\,\dd t_{1}
\\[3mm] = &\
{1 \over k}\,\Re\bracks{t - {\exp\pars{\ic k t} - 1 \over \ic k}} =
{1 \over k}\,\Re\bracks{t - {-\ic\cos\pars{kt} + \sin\pars{kt} + \ic \over k}} =
\color{#f00}{{kt - \sin\pars{kt} \over k^{2}}}
\end{align}
A: As I found, if you replace $t'_1$ and $t'_2$ with $t_1$ and $t_2$, respectively, and employ your method, then you will obtain the correct answer. Otherwise, as you have $t_1$ as upper limit of interior integral, it would be appeared in the final answer anyway (as it is seen in your answer), and so you cannot get the correct answer (the desired answer given by you).
