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: 26

Answer

$\frac{{316}}{{693}}$

Work Step by Step

$$\eqalign{ & \text{Let }I=\int_0^{\pi /4} {{{\sec }^6}\theta {{\tan }^6}\theta } d\theta \cr & {\text{Split }}{\sec ^6}\theta {\text{ as }}{\sec ^4}\theta {\sec ^2}\theta \cr &I = \int_0^{\pi /4} {{{\sec }^4}\theta {{\sec }^2}\theta {{\tan }^6}\theta } d\theta \cr & {\text{Use the property }}{\left( {{a^m}} \right)^n} = {a^{mn}} \cr & I = \int_0^{\pi /4} {{{\left( {{{\sec }^2}\theta } \right)}^2}{{\sec }^2}\theta {{\tan }^6}\theta } d\theta \cr & {\text{Rewrite }}{\sec ^2}\theta {\text{ using }}{\sec ^2}\theta = 1 + {\tan ^2}\theta \cr & I= \int_0^{\pi /4} {{{\left( {1 + {{\tan }^2}\theta } \right)}^2}{{\sec }^2}\theta {{\tan }^6}\theta } d\theta \cr & = \int_0^{\pi /4} {{{\left( {1 + {{\tan }^2}\theta } \right)}^2}{{\tan }^6}\theta {{\sec }^2}\theta } d\theta \cr & {\text{Expand the binomial}} \cr & I= \int_0^{\pi /4} {\left( {1 + 2{{\tan }^2}\theta + {{\tan }^4}\theta } \right){{\tan }^6}\theta {{\sec }^2}\theta } d\theta \cr & = \int_0^{\pi /4} {\left( {{{\tan }^6}\theta + 2{{\tan }^8}\theta + {{\tan }^{10}}\theta } \right){{\sec }^2}\theta } d\theta \cr & {\text{Integrate by the substitution method}} \cr & {\text{Let }}u = \tan \theta ,{\text{ }}du = {\sec ^2}\theta d\theta \cr & {\text{The new limits of integration are:}} \cr & x = \frac{\pi }{4} \to u = \tan \left( {\frac{\pi }{4}} \right) = 1 \cr & x = 0 \to u = \tan \left( 0 \right) = 0 \cr & {\text{Substituting}} \cr & I= \int_0^1 {\left( {{u^6} + 2{u^8} + {u^{10}}} \right)} du \cr & {\text{Use the power rule }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr & I= \left[ {\frac{{{u^7}}}{7} + \frac{{2{u^9}}}{9} + \frac{{{u^{11}}}}{{11}}} \right]_0^1 \cr & {\text{Evaluate}} \cr & I= \left[ {\frac{{{{\left( 1 \right)}^7}}}{7} + \frac{{2{{\left( 1 \right)}^9}}}{9} + \frac{{{{\left( 1 \right)}^{11}}}}{{11}}} \right] - \left[ {\frac{{{{\left( 0 \right)}^7}}}{7} + \frac{{2{{\left( 0 \right)}^9}}}{9} + \frac{{{{\left( 0 \right)}^{11}}}}{{11}}} \right] \cr & = \frac{1}{7} + \frac{2}{9} + \frac{1}{{11}} - 0 \cr & = \frac{{316}}{{693}} \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.