Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.4 Integration of Rational Functions by Partial Fractions - 7.4 Exercises - Page 542: 59

Answer

a)-c) See proof

Work Step by Step

a) If $t$ = $\tan(\frac{x}{2})$ then $\frac{x}{2}$ = $\tan^{-1}t$ $\cos\left(\frac{x}{2}\right)$ = $\frac{1}{\sqrt {1+t^2}}$ $\sin\left(\frac{x}{2}\right)$ = $\frac{t}{\sqrt {1+t^2}}$ b) $\cos x$ = $\cos\left(2\cdot\frac{x}{2}\right)$ = $2\cos^2\left(\frac{x}{2}\right)-1$ = $2\left(\frac{1}{\sqrt {1+t^2}}\right)^2-1$ = $\frac{1-t^2}{1+t^2}$ $\sin x =\sin\left(2\cdot\dfrac{x}{2}\right)= 2\sin\frac{x}{2}\cos \frac{x}{2}$ $ = 2\cdot\frac{t}{\sqrt{1+t^2}}\cdot\frac{1}{\sqrt{1+t^2}}=\frac{2t}{1+t^2}$ c) $\frac{x}{2}$ = $\tan^{-1}t$ $x$ = $2\tan^{-1}t$ $dx$ = $\frac{2}{1+t^2}dt$
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