Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Section 7.1 - Integration by Parts - 7.1 Exercises - Page 476: 22

Answer

$x(\arcsin x)^{2}+2\sqrt{1-x^{2}}\arcsin x-2x+C$

Work Step by Step

Integration by parts: $\displaystyle \int udv=uv-\int vdu$ The idea is to choose a relatively easy $dv$ to integrate, and a u whose $u'$ does not complicate matters (best case: makes thing simpler). ---- Let $u=(\arcsin x)^{2},\qquad dv=dx$ Then, $du=2\displaystyle \arcsin x\cdot\frac{1}{\sqrt{1-x^{2}}}, \qquad v=x$ $I=\displaystyle \int(\arcsin x)^{2}dx=uv-\int vdu$ $=x(\displaystyle \arcsin x)^{2}-2\int\arcsin x\cdot\frac{x}{\sqrt{1-x^{2}}}dx$ It might not look like it, but we have a "simpler" integrand. By parts again, $\left[\begin{array}{ll} u=\arcsin x & dv=\frac{x}{\sqrt{1-x^{2}}}\\ du=\frac{1}{\sqrt{1-x^{2}}} & v= -\sqrt{1-x^{2}} \end{array}\right],$ $($ for $\displaystyle \int\frac{xdx}{\sqrt{1-x^{2}}}$= substitute $\left[\begin{array}{l} t=1-x^{2}\\ dt=-2xdx \end{array}\right]$ = $-\displaystyle \frac{1}{2}\int t^{-1/2}dt=-\frac{1}{2}(\frac{t^{1/2}}{1/2})=-t^{1/2})$ $I=x(\displaystyle \arcsin x)^{2}-2(-\sqrt{1-x^{2}}\arcsin x-\int-1dx)$ $=x(\arcsin x)^{2}+2\sqrt{1-x^{2}}\arcsin x-2x+C$
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