University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Practice Exercises - Page 279: 120


$\sin^{-1} (\dfrac{\theta}{4})+C$

Work Step by Step

Calculate the anti-derivative. Then $\int \dfrac{d \theta}{\sqrt {16-\theta^2}}=\int \dfrac{d \theta}{\sqrt {4^2-\theta^2}}$ We know that $ \int \dfrac{dx}{\sqrt {a^2-x^2}} dx=\sin^{-1} (\dfrac{x}{a})+C$; where $C$ is a constant of proportionality. Thus, $= \sin^{-1} (\dfrac{\theta}{4})+C$
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