University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.3 - Trigonometric Substitutions - Exercises - Page 439: 8


$\dfrac{1}{6} \sin^{-1}(3t)+\dfrac{t\sqrt {1-9t^2}}{2}+C$

Work Step by Step

Let us consider $t =\dfrac{1}{3} \sin \theta \implies dt=\dfrac{1}{3} \cos \theta d \theta $ and $\cos \theta =\sqrt {1-9t^2}$ Now, the given integral becomes: $\int \sqrt{1-9(\dfrac{1}{9} \sin^2 \theta) } (\dfrac{1}{3} \cos \theta d \theta )=\dfrac{1}{3} \int (\cos \theta) (\cos \theta) d\theta $ As we know that $\cos^2 \theta=1-\sin^2 \theta$ and $\cos^2 \theta =(1/2)+(1/2) \cos 2 \theta$ $= \dfrac{1}{3} \int (\dfrac{1}{2}+\dfrac{1}{2} \cos ( 2\theta ))d \theta$ Now, integrate and plug in $\theta =\sin^{-1}(3t)$ and $t =\dfrac{1}{3} \sin \theta \implies \sin \theta =3t$ we have $ \dfrac{1}{3}(\dfrac{1}{2} \theta+\dfrac{1}{4} \sin ( 2\theta ))+C=\dfrac{1}{3}(\dfrac{1}{2} \theta+\dfrac{1}{4} (2 \sin \theta \cos \theta )))+C=\dfrac{1}{6} \sin^{-1}(3t)+\dfrac{1}{6}(3 t ) \sqrt {1-9t^2}+C $ or, $=\dfrac{1}{6} \sin^{-1}(3t)+\dfrac{t\sqrt {1-9t^2}}{2}+C$
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