Calculus 8th Edition

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

Chapter 6 - Inverse Functions - 6.4 Derivatives of Logarithmic Functions - 6.4 Exercises - Page 438: 79



Work Step by Step

Evaluate the integral $\int\frac{sin2x}{1+cos^{2}x}dx$ Consider $cos^{2}x=t$ and $2cosx(-sinx)dx=dt$ This implies $sin2xdx=-dt$ Thus, $\int\frac{sin2x}{1+cos^{2}x}dx=-\int\frac{1}{1+t}$ $=-ln|1+t|+constant$ Hence,$\int\frac{sin2x}{1+cos^{2}x}dx=-ln(1+cos^{2}x)+constant$
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