Answer
$$\ln (1+\sqrt{2}) \mathbf{i}+\left(1-\frac{\pi}{4}\right) \mathbf{j}+\left(\frac{\pi}{4 \sqrt{2}}-\frac{1}{\sqrt{2}}\right) \mathbf{k} $$
Work Step by Step
We integrate the vector function as follows:
\begin{align*}
\int_{0}^{\pi / 4}\left[(\sec t) \mathbf{i}+\left(\tan ^{2} t\right) \mathbf{j}-(t \sin t) \mathbf{k}\right] d t&=\int_{0}^{\pi / 4}\left[(\sec t) \mathbf{i}+\left(\sec ^{2} t-1\right) \mathbf{j}-(t \sin t) \mathbf{k}\right] d t \\
&=\ln |\sec t+\tan t|\bigg|_{0}^{\pi / 4} \mathbf{i}+(\tan t-t)\bigg|_{0}^{\pi / 4} \mathbf{j}+(t \cos t-\sin t)\bigg|_{0}^{\pi / 4} \mathbf{k}\\
&=\ln (1+\sqrt{2}) \mathbf{i}+\left(1-\frac{\pi}{4}\right) \mathbf{j}+\left(\frac{\pi}{4 \sqrt{2}}-\frac{1}{\sqrt{2}}\right) \mathbf{k}
\end{align*}
Where we used the fact that $\sec^2 x = \tan^2 x +1$
