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 346: 4


$$\int \sin^2\theta\cos\theta d\theta=\frac{\sin^3\theta}{3}+c$$

Work Step by Step

To solve the integral $\int \sin^2\theta\cos\theta d\theta$ we will use substitution $u=\sin\theta$ which gives us $du=\cos\theta d\theta.$ Putting this into our integral we get: $$\int \sin^2\theta\cos\theta d\theta=\int u^2du=\frac{u^3}{3}+c$$ where $c$ is arbitrary constant. Now we have to express solution in terms of $\theta$ by expressing $u$ in terms of $\theta$: $$\int \sin^2\theta\cos\theta d\theta=\frac{u^3}{3}+c=\frac{\sin^3\theta}{3}+c$$
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