Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises: 16

Answer

$\frac{\sin^3 x}{3}+C$

Work Step by Step

$\int \tan^2 x\cos^3 x\ dx$ $=\int\frac{\sin^2 x}{\cos^2 x}*\cos^3 x\ dx$ $=\int\sin^2 x\cos x\ dx$ We use u substitution. Let $u=\sin x$. Then, $du=\cos x\ dx$. $=\int u^2 du$ $=\frac{u^3}{3}+C$ $=\boxed{\frac{\sin^3 x}{3}+C}$
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