University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 337: 19


$$\int^{\pi}_05(5-4\cos t)^{1/4}\sin tdt=3^{5/2}-1$$

Work Step by Step

$$A=\int^{\pi}_05(5-4\cos t)^{1/4}\sin tdt$$ We set $u=5-4\cos t$, which means $$du=-4(-\sin t)dt=4\sin tdt$$ $$\sin tdt=\frac{1}{4}dt$$ For $t=\pi$, we have $u=5-4\cos\pi=5-4(-1)=9$ For $t=0$, we have $u=5-4\cos0=5-4(1)=1$ Therefore, $$A=\frac{5}{4}\int^9_1u^{1/4}du=\frac{5}{4}\times\frac{u^{5/4}}{\frac{5}{4}}\Big]^9_1=u^{5/4}\Big]^9_1$$ $$A=9^{5/4}-1$$ $$A=(3^2)^{5/4}-1$$ $$A=3^{5/2}-1$$
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.