Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.4 Trigonometric Substitutions - 7.4 Exercises: 14

Answer

$$18{\sin ^{ - 1}}\left( {\frac{t}{6}} \right) + \frac{1}{2}\sqrt {36 - {t^2}} + C$$

Work Step by Step

$$\eqalign{ & \int {\sqrt {36 - {t^2}} dt} \cr & {\text{substitute }}t = 6\sin \theta ,{\text{ }}dt = 6\cos \theta d\theta \cr & = \int {\sqrt {36 - 36{{\sin }^2}\theta } \left( {6\cos \theta } \right)d\theta } \cr & = \int {6\sqrt {1 - {{\sin }^2}\theta } \left( {6\cos \theta } \right)d\theta } \cr & {\text{pythagorean identity }}{\sin ^2}\theta + {\cos ^2}\theta = 1 \cr & = 36\int {\sqrt {{{\cos }^2}\theta } \cos \theta d\theta } \cr & = 36\int {{{\cos }^2}\theta d\theta } \cr & = 36\int {\left( {\frac{{1 + \cos 2\theta }}{2}} \right)} d\theta \cr & {\text{integrating}} \cr & = 36\left( {\frac{1}{2}\theta + \frac{1}{4}\sin 2\theta } \right) + C \cr & = 18\theta + 9\sin 2\theta \cr & = 18\theta + 18\sin \theta \cos \theta \cr & = 18{\sin ^{ - 1}}\left( {\frac{t}{6}} \right) + 18\left( {\frac{t}{6}} \right)\left( {\frac{{\sqrt {36 - {t^2}} }}{6}} \right) + C \cr & {\text{simplify}} \cr & = 18{\sin ^{ - 1}}\left( {\frac{t}{6}} \right) + \frac{1}{2}\sqrt {36 - {t^2}} + C \cr} $$
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.