Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 81

Answer

$$-\sqrt{7-t^2}+c .$$

Work Step by Step

Let $u=7-t^2$, then $du=-2tdt$, then $$\int \frac{t}{\sqrt{7-t^2}}dt=-\frac{1}{2}\int \frac{1}{\sqrt{u}}du=-\frac{1}{2}\frac{\sqrt u}{1/2}+c\\=-\sqrt{7-t^2}+c . $$
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