University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 320: 18

Answer

$4\sqrt{3}-3$

Work Step by Step

\begin{aligned} \int_{-\pi / 3}^{-\pi / 4}\left(4 \sec ^{2} t+\frac{\pi}{t^{2}}\right) d t&=\int_{-\pi / 3}^{-\pi / 4}\left(4 \sec ^{2} t+\pi t ^{-2}\right) d t\\ &=\left(4 \tan t-\pi t ^{-1}\right)\bigg|_{-\pi / 3}^{-\pi / 4}\\ &= 4(\tan(-\pi/4)-\tan(-\pi/3))-\pi(\frac{-4}{\pi}+\frac{3}{\pi})\\ &=4\left(-1+\sqrt{3}\right)+1\\ &=4\sqrt{3}-3 \end{aligned}
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