Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises - Page 499: 30

Answer

$\frac{\pi }{4} - \frac{2}{3}$

Work Step by Step

$$\eqalign{ & \text{Let }I=\int_0^{\pi /4} {{{\tan }^4}t} dt \cr & {\text{Split }}{\tan ^4}t{\text{ as }}{\tan ^2}t{\tan ^2}t \cr &I= \int_0^{\pi /4} {{{\tan }^2}t{{\tan }^2}t} dt \cr & {\text{Rewrite }}{\tan ^2}t{\text{ using }}{\sec ^2}\theta = 1 + {\tan ^2}\theta \cr & I = \int_0^{\pi /4} {\left( {{{\sec }^2}t - 1} \right){{\tan }^2}t} dt \cr & = \int_0^{\pi /4} {\left( {{{\sec }^2}t{{\tan }^2}t - {{\tan }^2}t} \right)} dt \cr & = \int_0^{\pi /4} {\left( {{{\tan }^2}t{{\sec }^2}t - {{\sec }^2}t + 1} \right)} dt \cr & {\text{Integrate by the substitution method}} \cr & {\text{Let }}u = \tan t,{\text{ }}du = {\sec ^2}tdt \cr & {\text{Substituting}} \cr & I= \int_0^{\pi /4} {\left( {{{\tan }^2}t{{\sec }^2}t - {{\sec }^2}t + 1} \right)} dt \cr & {\text{Integrating}} \cr & I= \left[ {\frac{{{{\tan }^3}t}}{3} - \tan t + t} \right]_0^{\pi /4} \cr & {\text{Evaluate the limits of integration}} \cr & I= \left[ {\frac{1}{3}{{\tan }^3}\left( {\frac{\pi }{4}} \right) - \tan \left( {\frac{\pi }{4}} \right) + \left( {\frac{\pi }{4}} \right)} \right] - \left[ {\frac{1}{3}{{\tan }^3}\left( 0 \right) - \tan \left( 0 \right) + \left( 0 \right)} \right] \cr & {\text{Simplifying}} \cr & I= \frac{1}{3}\left( 1 \right) - 1 + \frac{\pi }{4} - 0 \cr & = \frac{\pi }{4} - \frac{2}{3} \cr} $$
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