Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.2 Trigonometric Integrals - 7.2 Exercises: 22

Answer

$\frac{1}{3}\tan^3\theta+\frac{1}{5}\tan^5\theta+C$

Work Step by Step

$\int\tan^2\theta\sec^4\theta\ d\theta$ Since the power of secant is even, we save a factor of $\sec^2\theta$, and express the rest in terms of $\tan\theta$: $=\int\tan^2\theta\sec^2\theta\sec^2\theta\ d\theta$ $=\int\tan^2\theta(1+\tan^2\theta)\sec^2\theta\ d\theta$ Let $u=\tan\theta$. Then $du=\sec^2\theta\ d\theta$. $=\int u^2(1+u^2)du$ $=\int(u^2+u^4)du$ $=\frac{1}{3}u^3+\frac{1}{5}u^5+C$ $=\boxed{\frac{1}{3}\tan^3\theta+\frac{1}{5}\tan^5\theta+C}$
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