University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.2 - Trigonometric Integrals - Exercises - Page 434: 43


$$\int^{\pi/3}_{\pi/4}\tan^5\theta\sec^4\theta d\theta=\frac{43}{3}$$

Work Step by Step

$$A=\int^{\pi/3}_{\pi/4}\tan^5\theta\sec^4\theta d\theta$$ $$A=\int^{\pi/3}_{\pi/4}\tan^5\theta\sec^2\theta(\sec^2\theta d\theta)$$ $$A=\int^{\pi/3}_{\pi/4}\tan^5\theta\sec^2\theta d(\tan\theta)$$ We set $u=\tan\theta$, which means $\tan^5\theta=u^5$ Also, since $\sec^2\theta=\tan^2\theta+1$, we would have $\sec^2\theta=u^2+1$ - For $\theta=\pi/4$, $u=\tan(\pi/4)=1$ - For $\theta=\pi/3$, $u=\tan(\pi/3)=\sqrt3$ $$A=\int^{\sqrt3}_1u^5(u^2+1)du$$ $$A=\int^{\sqrt3}_1(u^7+u^5)du$$ $$A=\frac{u^8}{8}+\frac{u^6}{6}\Big]^{\sqrt3}_1$$ $$A=\frac{81}{8}+\frac{9}{2}-\Big(\frac{1}{8}+\frac{1}{6}\Big)$$ $$A=\frac{117}{8}-\frac{7}{24}=\frac{43}{3}$$
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.