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: 21


$$\int8\cos^32\theta\sin2\theta d\theta=2\sin^22\theta-\sin^42\theta+C=-\cos^4 2\theta+C$$

Work Step by Step

$$A=\int8\cos^32\theta\sin2\theta d\theta$$ $$A=\int4\cos^32\theta\sin2\theta d(2\theta)$$ We set $a=2\theta$. $$A=\int4\cos^3a\sin ada$$ We would rewrite $$\cos^3a=\cos^2a\cos a=(1-\sin^2a)\cos a$$ Therefore, $$A=\int4(1-\sin^2a)\sin a(\cos a da)$$ $$A=\int4(1-\sin^2a)\sin a d(\sin a)$$ Take $u=\sin a$ $$A=\int4(1-u^2)udu$$ $$A=4\int(u-u^3)du$$ $$A=4\Big(\frac{u^2}{2}-\frac{u^4}{4}\Big)+C$$ $$A=2u^2-u^4+C$$ $$A=2\sin^22\theta-\sin^42\theta+C$$ or (by using the trigonometric identity formula, $\sin^2+\cos^2=1$): $$A=-\cos^4 2\theta+C$$
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