Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.4 - Trigonometric Substitutions - Exercises 8.4 - Page 467: 8


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

Work Step by Step

Use formula: $\cos^2 \theta=1-\sin^2 \theta$ and $\cos^2 \theta =(1/2)+(1/2) \cos 2 \theta$ Plug $t =\dfrac{1}{3} \sin \theta \implies dt=\dfrac{1}{3} \cos \theta d \theta $ and $\cos \theta =\sqrt {1-9t^2}$ Then, $\int \sqrt{1-9(\dfrac{1}{9} \sin^2 \theta) } (\dfrac{1}{3} \cos \theta d \theta )=(\dfrac{1}{3}) \int (\cos \theta) (\cos \theta) d\theta =(\dfrac{1}{3}) \int (\dfrac{1}{2}+\dfrac{1}{2} \cos ( 2\theta ))d \theta$ Integrate and substitute in $\theta =\sin^{-1}(3t)$ and $\sin \theta =3t$. $ (\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)+(3t)(\dfrac{1}{6})(3\sqrt {1-9t^2}+C $ Thus, $\int \sqrt{1-9(\dfrac{1}{9} \sin^2 \theta) } (\dfrac{1}{3} \cos \theta d \theta )=\dfrac{1}{6} \sin^{-1}(3t)+\dfrac{t\sqrt {1-9t^2}}{2}+C$
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