Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises: 39


$$\int\frac{\sin2x}{1+\cos^2 x}dx=-\ln(1+\cos^2 x)+C$$

Work Step by Step

$$A=\int\frac{\sin2x}{1+\cos^2 x}dx$$ Let $u=1+\cos^2x $ Then apply the Chain Rule, we have $du=(1+\cos^2 x)'dx=2\cos x(\cos x)'dx=(-2\cos x\sin x)dx=-\sin2xdx$. That means $\sin2xdx=-du$ Substitute into $A$: $$A=-\int\frac{1}{u}du$$ $$A=-\ln|u|+C$$ $$A=-\ln|1+\cos^2x|+C$$ For all $x\in R$, $\cos^2x\geq0$, so it follows that $(1+\cos^2 x)\gt0$. Therefore, for all $x\in R$, $|1+\cos^2 x|=1+\cos^2 x$ So, $$A=-\ln(1+\cos^2 x)+C$$
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