Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Review - Exercises - Page 423: 38

Answer

$\int_{0}^{\pi/4} (1+tan~t)^3~sec^2~t~dt = \frac{15}{4}$

Work Step by Step

$\int_{0}^{\pi/4} (1+tan~t)^3~sec^2~t~dt$ Let $u = tan~t$ $\frac{du}{dt} = sec^2~t$ $dt = \frac{du}{sec^2~t}$ When $t = 0,~~$ then $~~u = 0$ When $t = \frac{\pi}{4},~~$ then $~~u = 1$ $\int_{0}^{1} (1+u)^3~sec^2~t~\frac{du}{sec^2~t}$ $= \int_{0}^{1} (1+u)^3~du$ $= \frac{1}{4}(1+u)^4~\Big\vert_{0}^{1}$ $= \frac{1}{4}(1+1)^4- \frac{1}{4}(1+0)^4$ $= \frac{1}{4}(16)- \frac{1}{4}(1)$ $= 4- \frac{1}{4}$ $= \frac{15}{4}$
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