Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Review - Exercises - Page 538: 34


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

Work Step by Step

$$ \int(\arcsin x)^{2} d x $$ Integrate by parts twice, first with $$ \left[\begin{array}{ll}{u=(\arcsin x)^{2},} & {d v=d x} \\ {d u=2\arcsin x \frac{d x}{\sqrt 1-x^{2}} } & {v= x}\end{array}\right] $$ $$ \int(\arcsin x)^{2} d x= x(\arcsin x)^{2}-\int 2 x \arcsin x\left(\frac{d x}{\sqrt{1-x^{2}}}\right) $$ Now let $$ \left[\begin{array}{ll}{U=\arcsin x ,} & {dV=\left(\frac{x d x}{\sqrt{1-x^{2}}}\right)} \\ {d U=\left(\frac{d x}{\sqrt{1-x^{2}}}\right)} & {V=-\sqrt{1-x^{2}} }\end{array}\right] $$ So $$ \begin{aligned} \int(\arcsin x)^{2} d x &=x(\arcsin x)^{2}-\int 2 x \arcsin x\left(\frac{d x}{\sqrt{1-x^{2}}}\right)\\ &=x(\arcsin x)^{2}-2\left[\arcsin x(-\sqrt{1-x^{2}})+\int d x\right]\\ &= x(\arcsin x)^{2}+2 \sqrt{1-x^{2}} \arcsin x-2 x+C \end{aligned} $$ where $C$ is constant.
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