Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises - Page 512: 9

Answer

$\int \frac{du}{\sqrt {a^{2}-u^{2}}}$ $u=t$ $a=1$

Work Step by Step

Try it: $u=t$ $du=dt$ $\int \frac{du}{\sqrt {a^{2}-u^{2}}}=arcsin\frac{u}{a}+C$ $3arcsint+C$ $\int \frac{du}{\sqrt {a^{2}-u^{2}}}$ $u=t$ $a=1$
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