91Ó°ÊÓ

The given definite integral contains two components: a component along the \(\mathbf{i}\) direction and a component along the \(\mathbf{j}\) direction. Separate the integral into these components: $$\int_{0}^{\ln 2}\left(e^{t} \mathbf{i}+e^{t} \cos \left(\pi e^{t}\right) \mathbf{j}\right) d t = \int_{0}^{\ln 2}e^{t} \mathbf{i} dt + \int_{0}^{\ln 2}e^{t} \cos \left(\pi e^{t}\right) \mathbf{j} dt$$

Evaluate the definite integral for the component along the \(\mathbf{i}\) direction: $$\int_{0}^{\ln 2}e^{t}\mathbf{i}dt = \left[e^{t}\mathbf{i}\right]_{0}^{\ln 2} = (e^{\ln 2}\mathbf{i} - e^{0}\mathbf{i}) = (2\mathbf{i} -\mathbf{i}) = \mathbf{i}$$

Evaluate the definite integral for the component along the \(\mathbf{j}\) direction: Since we have \(u=t\) and \(dv=e^{t} \cos \left(\pi e^{t}\right) dt\), we can find \(du=dt\) and \(v=-\frac{1}{\pi}\sin(\pi e^{t})\). Now, use integration by parts to solve this integral: $$\int_{0}^{\ln 2}e^{t} \cos \left(\pi e^{t}\right) \mathbf{j} dt = \left[-\frac{1}{\pi}\sin(\pi e^{t}) \cdot e^{t}\mathbf{j}\right]_{0}^{\ln 2} + \int_{0}^{\ln 2}\frac{1}{\pi}\sin(\pi e^{t})\mathbf{j}dt$$ Since \(\sin(0) = 0\) and \(\sin(\pi) = 0\), the first term evaluates to zero. Now evaluate the remaining integral: $$\int_{0}^{\ln 2}\frac{1}{\pi}\sin(\pi e^{t})\mathbf{j}dt = \left[-\frac{1}{\pi^2} \cos(\pi e^{t})\mathbf{j}\right]_{0}^{\ln 2} = \left(-\frac{1}{\pi^2}\cos(\pi) - (-\frac{1}{\pi^2}\cos(0))\right)\mathbf{j} = \frac{2}{\pi^2}\mathbf{j}$$

Combine the results from steps 2 and 3 to obtain the evaluated definite integral: $$\int_{0}^{\ln 2}\left(e^{t} \mathbf{i}+e^{t} \cos \left(\pi e^{t}\right) \mathbf{j}\right) d t = \mathbf{i} + \frac{2}{\pi^2}\mathbf{j}$$