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 - Page 484: 2


$$\int\sin^3\theta\cos^4\theta d\theta=\frac{\cos^7\theta}{7}-\frac{\cos^5\theta}{5}+C$$

Work Step by Step

$$A=\int\sin^3\theta\cos^4\theta d\theta$$ Since the power of $\sin$ is odd, the strategy is to save 1 $\sin$ factor and change the rest in terms of $\cos$ according to the formula $\sin^2\theta=1-\cos^2\theta$ Therefore, we have $$A=\int\sin^2\theta\cos^4\theta\sin \theta d\theta$$ $$A=\int\cos^4\theta(1-\cos^2\theta)\sin\theta d\theta$$ Now we substitute $u=\cos\theta$ That makes $du=-\sin\theta d\theta$, or $\sin\theta d\theta=-du$ So, $$A=-\int u^4(1-u^2)du$$ $$A=-\int (u^4-u^6)du$$ $$A=\int (u^6-u^4)du$$ $$A=\frac{u^7}{7}-\frac{u^5}{5}+C$$ $$A=\frac{\cos^7\theta}{7}-\frac{\cos^5\theta}{5}+C$$
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