Answer
$$\frac{{{e^{ - t}}}}{2}\left( {\cos t - \sin t} \right){\bf{i}} + \frac{{{e^{ - t}}}}{2}\left( {\sin t - \cos t} \right){\bf{j}} + {\bf{C}}$$
Work Step by Step
$$\eqalign{
& \int {\left( {{e^{ - t}}\sin t{\bf{i}} + {e^{ - t}}\cos t{\bf{j}}} \right)} dt \cr
& {\text{ By the Definition of Integration of Vector - Valued Functions}} \cr
& = \left[ {\int {{e^{ - t}}\sin tdt} } \right]{\bf{i}} + \left[ {\int {{e^{ - t}}\cos tdt} } \right]{\bf{j}} \cr
& {\text{By the tables of integration}} \cr
& \int {{e^{au}}\sin bu} du = \frac{{{e^{au}}}}{{{a^2} + {b^2}}}\left( {a\sin bu - b\cos bu} \right) + C \cr
& \int {{e^{ - t}}\sin u} du = \frac{{{e^{ - t}}}}{{{{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2}}}\left( { - \sin t - \cos t} \right) + C \cr
& \int {{e^{ - t}}\sin u} du = \frac{{{e^{ - t}}}}{2}\left( {\cos t - \sin t} \right) + C \cr
& and \cr
& \int {{e^{au}}\cos bu} du = \frac{{{e^{au}}}}{{{a^2} + {b^2}}}\left( {a\cos bu + b\sin bu} \right) + C \cr
& \int {{e^{ - t}}\cos tdt} = \frac{{{e^{ - t}}}}{{{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}}}\left( { - \cos t + \sin t} \right) + C \cr
& \int {{e^{ - t}}\cos tdt} = \frac{{{e^{ - t}}}}{2}\left( {\sin t - \cos t} \right) + C \cr
& {\text{Therefore,}} \cr
& = \left[ {\int {{e^{ - t}}\sin tdt} } \right]{\bf{i}} + \left[ {\int {{e^{ - t}}\cos tdt} } \right]{\bf{j}} \cr
& = \frac{{{e^{ - t}}}}{2}\left( {\cos t - \sin t} \right){\bf{i}} + \frac{{{e^{ - t}}}}{2}\left( {\sin t - \cos t} \right){\bf{j}} + {\bf{C}} \cr
& {\text{where }}{\bf{C}}{\text{ is a constant vector}} \cr} $$
