Calculus: Early Transcendentals (2nd Edition)

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

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises: 49

Answer

$$6{\sin ^{ - 1}}\frac{x}{5} + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{6}{{\sqrt {25 - {x^2}} }}dx} \cr & {\text{take out the constant}} \cr & = 6\int {\frac{1}{{\sqrt {25 - {x^2}} }}dx} \cr & {\text{from the table 4}}{\text{.10 }}\int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }}} = {\sin ^{ - 1}}\frac{x}{a} + C \cr & {\text{letting }}a = 5 \cr & = 6{\sin ^{ - 1}}\frac{x}{5} + C \cr & {\text{check by differentiation}} \cr & {\text{ = }}\frac{d}{{dx}}\left( {6{{\sin }^{ - 1}}\frac{x}{5} + C} \right) \cr & {\text{ = 6}}\frac{d}{{dx}}\left( {{{\sin }^{ - 1}}\frac{x}{5}} \right) + \frac{d}{{dx}}\left( C \right) \cr & {\text{ = }}6\left( {\frac{1}{{\sqrt {{{\left( 5 \right)}^2} - {x^2}} }}} \right) + 0 \cr & = \frac{6}{{\sqrt {25 - {x^2}} }} \cr} $$
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.