Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 11 - Vectors and Vector-Valued Functions - 11.6 Calculus of Vector-Valued Functions - 11.6 Exercises - Page 815: 62

Answer

\[\frac{3}{2}\ln \left( \frac{3}{2} \right)\mathbf{i}-2\mathbf{k}\]

Work Step by Step

\[\begin{align} & \int_{1/2}^{1}{\left( \frac{3}{1+2t}\mathbf{i}-\pi {{\csc }^{2}}\left( \frac{\pi }{2}t \right)\mathbf{k} \right)dt} \\ & \text{Integrate} \\ & =\left[ \frac{3}{2}\ln \left| 1+2t \right|\mathbf{i}+\pi \left( \frac{2}{\pi } \right)\cot \left( \frac{\pi }{2}t \right)\mathbf{k} \right]_{1/2}^{1} \\ & =\left[ \frac{3}{2}\ln \left| 1+2t \right|\mathbf{i}+2\cot \left( \frac{\pi }{2}t \right)\mathbf{k} \right]_{1/2}^{1} \\ & \text{Evaluate} \\ & =\left[ \frac{3}{2}\ln \left| 1+2\left( 1 \right) \right|\mathbf{i}+2\cot \left( \frac{\pi }{2}\left( 1 \right) \right)\mathbf{k} \right] \\ & -\left[ \frac{3}{2}\ln \left| 1+2\left( \frac{1}{2} \right) \right|\mathbf{i}+2\cot \left( \frac{\pi }{2}\left( \frac{1}{2} \right) \right)\mathbf{k} \right] \\ & \text{Simplifying} \\ & =\frac{3}{2}\ln \left| 3 \right|\mathbf{i}+2\cot \left( \frac{\pi }{2} \right)\mathbf{k}-\frac{3}{2}\ln \left| 2 \right|\mathbf{i}-2\cot \left( \frac{\pi }{4} \right)\mathbf{k} \\ & =\frac{3}{2}\ln 3\mathbf{i}+2\left( 0 \right)\mathbf{k}-\frac{3}{2}\ln 2\mathbf{i}-2\left( 1 \right)\mathbf{k} \\ & =\frac{3}{2}\ln 3\mathbf{i}-\frac{3}{2}\ln 2\mathbf{i}-2\mathbf{k} \\ & =\frac{3}{2}\ln \left( \frac{3}{2} \right)\mathbf{i}-2\mathbf{k} \\ \end{align}\]
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