Answer
$$\int_{0}^{\pi / 4}[(\sec t \tan t) \mathbf{i}+(\tan t) \mathbf{j}+(2 \sin t \cos t) \mathbf{k}] d t=(\sqrt{2}-1) \mathbf{i}+\ln \sqrt{2} \ \mathbf{j}+\frac{1}{2} \mathbf{k}$$
Work Step by Step
Given $$\int_{0}^{\pi / 4}[(\sec t \tan t) \mathbf{i}+(\tan t) \mathbf{j}+(2 \sin t \cos t) \mathbf{k}] d t$$ Integrating on a component-by-component basis produces \begin{align} I&=\int_{0}^{\pi / 4}[(\sec t \tan t) \mathbf{i}+(\tan t) \mathbf{j}+(2 \sin t \cos t) \mathbf{k}] d t\\ &=\left[\sec t \mathbf{i}+\ln |\sec t| \mathbf{j}+\sin ^{2} t \mathbf{k}\right]_{0}^{\pi / 4}\\ &=(\sec \frac{\pi}{4}-\sec0) \mathbf{i}+(\ln \sec \frac{\pi}{4}-\ln \sec 0) \mathbf{j}+(\sin^2 \frac{\pi}{4}-\sin^20 ) \ \mathbf{k} \\ &=(\sqrt{2}-1) \mathbf{i}+\ln \sqrt{2} \mathbf{j}+\frac{1}{2} \mathbf{k} \\ \end{align}
