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: 35


$$\int\sqrt{\cot x}\csc^2xdx=-\frac{2}{3}\sqrt{\cot^3x}+C$$

Work Step by Step

$$A=\int\sqrt{\cot x}\csc^2xdx$$ Let $u=\cot x$ Then we have $du=-\csc^2 xdx$. That means $\csc^2dx=-du$ Also, $\sqrt{\cot x}=\sqrt u=u^{1/2}$ Substitute into $A$: $$A=-\int u^{1/2}du$$ $$A=-\frac{u^{3/2}}{\frac{3}{2}}+C$$ $$A=-\frac{2\sqrt{u^3}}{3}+C$$ $$A=-\frac{2}{3}\sqrt{\cot^3x}+C$$
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