Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.5 The Substitution Rule - 4.5 Exercises: 7

Answer

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

Work Step by Step

To solve the integral $\int x\sqrt{1-x^2}dx$ we will use substitution $u=1-x^2$ which gives us $du=-2xdx\Rightarrow-\frac{1}{2}du=xdx$. Putting this in the integral we get: $$\int x\sqrt{1-x^2}dx=\int\sqrt u\cdot(-\frac{1}{2}du)=-\frac{1}{2}\int u^{1/2}du=-\frac{1}{2}\frac{u^{3/2}}{\frac{3}{2}}+c=-\frac{1}{2}\frac{2}{3}u^{3/2}+c=-\frac{1}{3}u^{3/2}+c,$$ where $c$ is arbitrary constant. Now we have to express solution in terms of $x$ by expressing $u$ in terms of $x$: $$\int x\sqrt{1-x^2}dx=-\frac{1}{3}u^{3/2}+c=-\frac{1}{3}(1-x^2)^{3/2}+c$$
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