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: 7

Answer

$$\int x\sqrt{1-x^2}dx=-\frac{1}{3}\sqrt{(1-x^2)^3}+C$$

Work Step by Step

$$A=\int x\sqrt{1-x^2}dx$$ Let $u=1-x^2$ Then $du=(1-x^2)'dx=-2xdx$. So $xdx=-\frac{1}{2}du$ Substitute into $A$, we have $$A=\int \sqrt u(-\frac{1}{2})du$$ $$A=-\frac{1}{2}\int u^{1/2}du$$ $$A=-\frac{1}{2}\frac{u^{3/2}}{\frac{3}{2}}+C$$ $$A=-\frac{\sqrt{u^3}}{3}+C$$ $$A=-\frac{1}{3}\sqrt{(1-x^2)^3}+C$$
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