University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.5 - Integral Tables and Computer Algebra Systems - Exercises - Page 451: 51


$$\int e^t\sec^3(e^t-1)dt=\frac{\sec (e^t-1)\tan (e^t-1)}{2}+\frac{1}{2}\ln|\sec(e^t-1)+\tan(e^t-1)|+C$$

Work Step by Step

$$A=\int e^t\sec^3(e^t-1)dt$$ Set $u=e^t-1$, which means $$du=e^tdt$$ Therefore, $$A=\int \sec^3udu$$ Use Reduction Formula 99, which states that $$\int \sec^naxdx=\frac{\sec^{n-2}ax\tan ax}{a(n-1)}+\frac{n-2}{n-1}\int\sec^{n-2}axdx$$ for $n=3$ and $a=1$: $$A=\frac{\sec u\tan u}{2}+\frac{1}{2}\int\sec udu$$ $$A=\frac{\sec u\tan u}{2}+\frac{1}{2}\ln|\sec u+\tan u|+C$$ $$A=\frac{\sec (e^t-1)\tan (e^t-1)}{2}+\frac{1}{2}\ln|\sec(e^t-1)+\tan(e^t-1)|+C$$
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