Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - Review Exercises: 17

Answer

$\frac{1}{\pi}\sin(\pi x-1)-\frac{1}{3\pi}\sin^{3}(\pi x-1)+C$

Work Step by Step

$\int \cos^3(\pi x-1)dx$ $=\int \cos^2(\pi x-1)\cos(\pi x-1)dx$ $=\int (1-\sin^2(\pi x-1))(\cos(\pi x-1))dx$ $=\int (\cos(\pi x-1)dx -\int\sin^2(\pi x-1)\cos(\pi x-1)dx$ Use u-substitution for both parts (they have different u's) Let $u= \pi x-1$ $\frac{du}{dx}= \pi$ $dx=\frac{1}{\pi}du$ Let $u=\sin(\pi x-1)$ $\frac{du}{dx}=\pi \cos(\pi x -1)$ $dx=\frac{1}{\pi \cos(\pi x -1)}du$ $=\int \frac{1}{\pi}\cos udu-\int \frac{1}{\pi}u^{2}du$ $=\frac{1}{\pi}\sin(\pi x-1)-\frac{1}{3\pi}\sin^{3}(\pi x-1)+C$
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