Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.6 - Integral Tables and Computer Algebra Systems - Exercises 8.6 - Page 481: 34

Answer

$-\dfrac{1}{2}\ln \space|\dfrac{2+\sqrt{4-\sin^2 (t)}}{\sin t}|+C$

Work Step by Step

Let $I=\int\dfrac{\cos t \space dt}{\sin t\sqrt{4-\sin^2t}}$ Suppose $u=\sin t \implies du=\cos t \space dt$ Therefore, $I=\int\dfrac{du}{u\sqrt{4-u^2}}du \space \\ =-\dfrac{1}{2}\ln|\dfrac{2+\sqrt{4-u^2}}{u}|+C \\=-\dfrac{1}{2}\ln \space|\dfrac{2+\sqrt{4-\sin^2 (t)}}{\sin t}|+C$
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