University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Practice Exercises - Page 343: 43


$$\int2(\cos x)^{-1/2}\sin xdx=-4\sqrt{\cos x}+C$$

Work Step by Step

$$A=\int2(\cos x)^{-1/2}\sin xdx$$ We set $a=\cos x$, then $$da=-\sin xdx$$ $$\sin xdx=-da$$ Therefore, $$A=-\int2a^{-1/2}da$$ $$A=-\frac{2a^{1/2}}{\frac{1}{2}}+C$$ $$A=-4a^{1/2}+C=-4\sqrt a+C$$ $$A=-4\sqrt{\cos x}+C$$
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