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: 27


$$\int^{\pi}_{0}\frac{\sin t}{2-\cos t}dt=\ln3$$

Work Step by Step

$$A=\int^{\pi}_{0}\frac{\sin t}{2-\cos t}dt$$ We set $u=2-\cos t$, which means $$du=-(-\sin t)dt=\sin tdt$$ For $t=\pi$, we have $$u=2-\cos\pi=2-(-1)=3$$ For $t=0$, we have $$u=2-\cos0=2-1=1$$ Therefore, $$A=\int^3_1\frac{1}{u}du=\ln|u|\Big]^3_1$$ $$A=\ln3-\ln1=\ln3-0$$ $$A=\ln3$$
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.