University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Practice Exercises - Page 343: 64


$$\int \frac{6dr}{\sqrt{4-(r+1)^2}}=6\sin^{-1}\Big(\frac{r+1}{2}\Big)+C$$

Work Step by Step

$$A=\int \frac{6dr}{\sqrt{4-(r+1)^2}}$$ $$A=6\int\frac{dr}{\sqrt{4-(r+1)^2}}$$ We set $a=r+1$, which means $$da=dr$$ Therefore, $$A=6\int\frac{da}{\sqrt{4-a^2}}$$ We have $$\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\Big(\frac{x}{a}\Big)+C$$ So, $$A=6\sin^{-1}\Big(\frac{a}{2}\Big)+C$$ $$A=6\sin^{-1}\Big(\frac{r+1}{2}\Big)+C$$
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.