Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.3 Trigonometric Substitution - Exercises - Page 410: 4

Answer

see below answer

Work Step by Step

a) $x$ = $2tanθ$ $dx$ = $2sec^{2}θdθ$ $I$ = $\int{\frac{dx}{(x^{2}+4)^{2}}}$ $I$ = $\int{\frac{2sec^{2}θdθ}{(4tan^{2}θ+4)^{2}}}$ $I$ = $\frac{1}{8}\int{\frac{sec^{2}θdθ}{(sec^{2}θ)^{2}}}$ $I$ = $\frac{1}{8}\int{cos^{2}θ}dθ$ b) $I$ = $\frac{1}{8}\int{cos^{2}θ}dθ$ $I$ = $\frac{1}{8}\int{\frac{1+cos2θ}{2}}dθ$ $I$ = $\frac{1}{16}\int{dθ}+\frac{1}{8}\int{cos2θ}dθ$ $I$ = $\frac{1}{16}θ+\frac{1}{16}sin2θ+C$ $I$ = $\frac{1}{16}θ+\frac{1}{16}sinθcosθ+C$ c) $x$ = $2tanθ$ $tanθ$ = $\frac{x}{2}$ $sinθ$ =$\frac{x}{\sqrt {x^{2}+4}}$ $cosθ$ =$\frac{2}{\sqrt {x^{2}+4}}$ d) $I$ = $\frac{1}{16}θ+\frac{1}{16}sinθcosθ+C$ $I$ = $\frac{1}{16}tan^{-1}(\frac{x}{2})+\frac{x}{8(x^{2}+4)}+C$
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