#### Answer

$${\bf{i}}$$

#### Work Step by Step

$$\eqalign{
& \int_0^{\ln 2} {\left( {{e^t}{\bf{i}} + {e^t}\cos \left( {\pi {e^t}} \right){\bf{j}}} \right)} dt \cr
& {\text{use }}\int_a^b {{\bf{r}}\left( t \right)} dt = \left( {\int_a^b {f\left( t \right)} dt} \right){\bf{i}} + \left( {\int_a^b {g\left( t \right)} dt} \right){\bf{j}} + \left( {\int_a^b {h\left( t \right)} dt} \right){\bf{k}}{\text{ }}\left( {{\text{see page 814}}} \right) \cr
& {\text{ then}} \cr
& = \left( {\int_0^{\ln 2} {{e^t}} dt} \right){\bf{i}} + \left( {\int_0^{\ln 2} {{e^t}\cos \left( {\pi {e^t}} \right)} dt} \right){\bf{j}} \cr
& = \left( {\int_0^{\ln 2} {{e^t}} dt} \right){\bf{i}} + \left( {\frac{1}{\pi }\int_0^{\ln 2} {\cos \left( {\pi {e^t}} \right)\left( {{e^t}\pi } \right)} dt} \right){\bf{j}} \cr
& {\text{evaluate integrals for each component}} \cr
& = \left( {{e^t}} \right)_0^{\ln 2}{\bf{i}} + \left( {\sin \left( {\pi {e^t}} \right)} \right)_0^{\ln 2}{\bf{j}} \cr
& = \left( {{e^{\ln 2}} - {e^0}} \right){\bf{i}} + \left( {\sin \left( {\pi {e^{\ln 2}}} \right) - \sin \left( {\pi {e^0}} \right)} \right){\bf{j}} \cr
& {\text{simplifying}} \cr
& = \left( {2 - 1} \right){\bf{i}} + \left( {\sin \left( {2\pi } \right) - \sin \left( \pi \right)} \right){\bf{j}} \cr
& = {\bf{i}} + \left( 0 \right){\bf{j}} \cr
& = {\bf{i}} \cr} $$