Calculus 8th Edition

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

Chapter 4 - Integrals - 4.5 The Substitution Rule - 4.5 Exercises - Page 347: 77



Work Step by Step

To evaluate the integral $$\int\frac{\sin2x}{1+\cos^2x}dx$$ we will use substitution $1+\cos^2x=t$ which gives us $dt=(1+\cos^2x)'dx=-2\cos x\sin xdx=-\sin2xdx\Rightarrow -dt=\sin2xdx$. Putting this into the integral we get: $$\int\frac{\sin2x}{1+\cos^2x}dx=\int\frac{1}{t}(-dt)=-\ln|t|+c$$ where $c$ is arbitrary constant. Now we have to express solution in terms of $x$: $$\int\frac{\sin2x}{1+\cos^2x}dx=-\ln|t|+c=-\ln(1+\cos^2x)+c$$
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