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: 40

Answer

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

Work Step by Step

$$A=\int\frac{\sin x}{1+\cos^2 x}dx$$ Let $u=\cos x$ Then we have $du=-\sin x'dx$. That means $\sin x dx=-du$ Substitute into $A$: $$A=\int-\frac{1}{1+u^2}du$$ The differentiation of $\cot^{-1}x$ is $-\frac{1}{1+x^2}$. Therefore, since $A=\int-\frac{1}{1+u^2}du$, $$A=\cot^{-1}u+C$$ $$A=\cot^{-1}(\cos x)+C$$
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