Answer
$$\int_{0}^{\pi / 2}[(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}+\mathbf{k}] d t=a \mathbf{i}+a \mathbf{j}+\frac{\pi}{2} \mathbf{k}$$
Work Step by Step
Given $$\int_{0}^{\pi / 2}[(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}+\mathbf{k}] d t $$
Integrating on a component-by-component basis produces
\begin{align}
I&=\int_{0}^{\pi / 2}[(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}+\mathbf{k}] d t\\
&=[a \sin t \mathbf{i}]_{0}^{\pi / 2}-[a \cos t \mathbf{j}]_{0}^{\pi / 2}+[t\mathbf{k}]_{0}^{\pi / 2}\\
&= a( \sin \frac{\pi}{2}-\sin 0) \mathbf{i} - a (\cos\frac{\pi}{2}-\cos0) \mathbf{j} + (\frac{\pi}{2}-0) \mathbf{k}] \\
&=a \mathbf{i}+a \mathbf{j}+\frac{\pi}{2} \mathbf{k}
\end{align}
