Answer
$$\mathbf{i}+(\ln 2) \mathbf{j}+\frac{3}{4} \mathbf{k} $$
Work Step by Step
Given $$ \int_{0}^{\pi / 3}[(\sec t \tan t) \mathbf{i}+(\tan t) \mathbf{j}+(2 \sin t \cos t) \mathbf{k}] d t $$
Then
\begin{align*}
\int_{0}^{\pi / 3}[(\sec t \tan t) \mathbf{i}+(\tan t) \mathbf{j}+(2 \sin t \cos t) \mathbf{k}] d t&=\int_{0}^{\pi / 3}[(\sec t \tan t) \mathbf{i}+(\tan t) \mathbf{j}+(\sin 2 t) \mathbf{k}] d t\\
&=[\sec t]\bigg|_{0}^{\pi / 3} \mathbf{i}+[-\ln (\cos t)]\bigg|_{0}^{\pi / 3} \mathbf{j}+\left[-\frac{1}{2} \cos 2 t\right]\bigg|_{0}^{\pi / 3} \mathbf{k}\\
&=\mathbf{i}+(\ln 2) \mathbf{j}+\frac{3}{4} \mathbf{k}
\end{align*}
