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: 78


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

Work Step by Step

To evaluate the integral $$\int\frac{\sin x}{1+\cos^2x}dx$$ we will use substitution $\cos x=t$ which gives us $-\sin xdx=dt\Rightarrow \sin xdx=-dt$. Putting this into the integral we get: $$\int\frac{\sin x}{1+\cos^2x}dx=\int\frac{1}{1+t^2}(-dt)=-\arctan t+c$$ where $c$ is arbitrary constant. Now we have to express solution in terms of $x$: $$\int\frac{\sin x}{1+\cos^2x}dx=-\arctan t+c=-\arctan(\cos x)+c$$
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