Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.2 - Integrals of Vector Functions; Projectile Motion - Exercises 13.2 - Page 753: 10


$$\ln (1+\sqrt{2}) \mathbf{i}+\left(1-\frac{\pi}{4}\right) \mathbf{j}+\left(\frac{\pi}{4 \sqrt{2}}-\frac{1}{\sqrt{2}}\right) \mathbf{k} $$

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 $\sec^2 x = \tan^2 x +1$
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