Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises - Page 418: 10


$$\int\sin t\sqrt{1+\cos t}dt=-\frac{2}{3}\sqrt{(1+\cos t)^3}+C$$

Work Step by Step

$$A=\int\sin t\sqrt{1+\cos t}dt$$ Let $u=1+\cos t$ Then $dt=(1+\cos t)'dt=-\sin tdt$. So $\sin tdt=-du$ Also, $\sqrt{1+\cos t}=\sqrt u=u^{1/2}$ Substitute into $A$, we have $$A=-\int u^{1/2}du$$ $$A=-\frac{u^{3/2}}{\frac{3}{2}}+C$$ $$A=-\frac{2\sqrt{u^3}}{3}+C$$ $$A=-\frac{2}{3}\sqrt{(1+\cos t)^3}+C$$
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