Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises - Page 484: 26

Answer

$\frac{316}{693}$

Work Step by Step

$\int_0^{\pi/4}\sec^6\theta\tan^6\theta\ d\theta$ Since the power of secant is even, save a factor of $\sec^2 \theta$ and express the rest in terms of $\tan \theta$. $=\int_0^{\pi/4}\sec^4\theta\tan^6\theta\sec^2\theta\ d\theta$ $=\int_0^{\pi/4}(\sec^2\theta)^2\tan^6\theta\sec^2\theta\ d\theta$ $=\int_0^{\pi/4}(1+\tan^2\theta)^2\tan^6\theta\sec^2\theta\ d\theta$ Let $u=\tan\theta$. Then $du=\sec^2\theta\ d\theta$. $=\int_{\tan 0}^{\tan(\pi/4)}(1+u^2)^2 u^6\ du$ $=\int_0^1(1+2u^2+u^4)u^6\ du$ $=\int_0^1(u^{10}+2u^8+u^6)\ du$ $=(\frac{u^{11}}{11}+\frac{2u^9}{9}+\frac{u^7}{7})|_0^1$ $=(\frac{1^{11}}{11}+\frac{2*1^9}{9}+\frac{1^7}{7})-(\frac{0^{11}}{11}+\frac{2*0^9}{9}+\frac{0^7}{7})$ $=(\frac{1}{11}+\frac{2}{9}+\frac{1}{7})-(0+0+0)$ $=\frac{63}{693}+\frac{154}{693}+\frac{99}{693}$ $=\boxed{\frac{316}{693}}$
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