University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 427: 12


$$\int \sin^{-1}ydy=y\sin^{-1}y+\sqrt{1-y^2}+C$$

Work Step by Step

$$A=\int \sin^{-1}ydy$$ Set $u=\sin^{-1}y$ and $dv=dy$ Then we would have $du=\frac{1}{\sqrt{1-y^2}}dy$ and $v=y$ Using the formula $\int udv= uv-\int vdu$: $$A=y\sin^{-1}y-\int\frac{y}{\sqrt{1-y^2}}dy$$ Set $a=\sqrt{1-y^2}$ Then we would have $$da=\frac{(1-y^2)'}{2\sqrt{1-y^2}}dy=\frac{-2y}{2\sqrt{1-y^2}}dy=-\frac{y}{\sqrt{1-y^2}}dy$$ Therefore, $$\frac{y}{\sqrt{1-y^2}}dy=-da$$ Substitute these back into $A$: $$A=y\sin^{-1}y-\int-da$$ $$A=y\sin^{-1}y+a+C$$ $$A=y\sin^{-1}y+\sqrt{1-y^2}+C$$
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