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 419: 38


$$\int\frac{dt}{\cos^2 t\sqrt{1+\tan t}}=2\sqrt{1+\tan t}+C$$

Work Step by Step

$$A=\int\frac{dt}{\cos^2 t\sqrt{1+\tan t}}$$ Let $u=1+\tan t $ Then we have $du=\frac{1}{\cos^2 t}dt$. Also, $\sqrt{1+\tan t}=\sqrt u=u^{1/2}$ Substitute into $A$: $$A=\int\frac{1}{u^{1/2}}du$$ $$A=\int(u^{-1/2})du$$ $$A=\frac{u^{1/2}}{\frac{1}{2}}+C$$ $$A=2\sqrt u+C$$ $$A=2\sqrt{1+\tan t}+C$$
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