University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.2 - Trigonometric Integrals - Exercises - Page 434: 16


$$\int7\cos^7tdt=7\sin t-7\sin^3 t+\frac{21\sin^5t}{5}-\sin^7t+C$$

Work Step by Step

$$A=\int7\cos^7tdt$$ This is Case 2, so we need to find a way to rewrite $$(\cos^2x)^k\cos x=(1-\sin^2x)^k\cos x$$ We have, $$A=\int7(\cos^2t)^3\cos tdt$$ $$A=7\int(1-\sin^2t)^3d(\sin t)$$ We would take $u=\sin t$ $$A=7\int(1-u^2)^3du$$ $$A=7\int(1-3u^2+3u^4-u^6)du$$ $$A=7\Big(u-u^3+\frac{3u^5}{5}-\frac{u^7}{7}\Big)+C$$ $$A=7\sin t-7\sin^3 t+\frac{21\sin^5t}{5}-\sin^7t+C$$
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