Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Section 7.6 - Integration Using Tables and Computer Algebra Systems - 7.6 Exercises - Page 513: 24


$\displaystyle \frac{2x^{4}-1}{8}\sin^{-1}(x^{2})+\frac{x^{2}\sqrt{1-x^{4}}}{8}+C$

Work Step by Step

$\displaystyle \int x^{3}\arcsin(x^{2})dx=\quad$substitute $\left[\begin{array}{ll} u=x^{2} & \\ du=2xdx & xdx=\frac{1}{2}du \end{array}\right]$ $=\displaystyle \frac{1}{2}\int u\arcsin(u)du=$ Table of integrals: $\color{blue}{ 90. \quad \displaystyle \int u\sin^{-1}udu=\frac{2u^{2}-1}{4}\sin^{-1}u+\frac{u\sqrt{1-u^{2}}}{4}+C }$ $=\displaystyle \frac{2u^{2}-1}{8}\sin^{-1}u+\frac{u\sqrt{1-u^{2}}}{8}+C$ ... bring back x... $=\displaystyle \frac{2x^{4}-1}{8}\sin^{-1}(x^{2})+\frac{x^{2}\sqrt{1-x^{4}}}{8}+C$
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