Answer
$$\mathbf{i}-\mathbf{j}+\frac{\pi}{4} \mathbf{k}$$
Work Step by Step
We evaluate the integral of the vector function as follows:
\begin{align*}
\int_{0}^{\pi / 2}\left[(\cos t) \mathbf{i}-(\sin 2 t) \mathbf{j}+\left(\sin ^{2} t\right) \mathbf{k}\right] d t&=\int_{0}^{\pi / 2}\left[(\cos t) \mathbf{i}-(\sin 2 t) \mathbf{j}+\left(\frac{1}{2}-\frac{1}{2} \cos 2 t\right) \mathbf{k}\right] d t \\
&=\sin t\bigg|_{0}^{\pi / 2} \mathbf{i}+ \frac{1}{2} \cos t\bigg|_{0}^{\pi / 2} \mathbf{j}+\left(\frac{1}{2} t-\frac{1}{4} \sin 2 t\right)\bigg|_{0}^{\pi / 2} \mathbf{k}\\
&=\mathbf{i}-\mathbf{j}+\frac{\pi}{4} \mathbf{k}
\end{align*}
