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: 13

Answer

$$\int\sqrt{\cos\theta}\sin^3\theta d\theta=\frac{2(\cos\theta)^{7/2}}{7}-\frac{2(\cos\theta)^{3/2}}{3}+C$$

Work Step by Step

$$A=\int\sqrt{\cos\theta}\sin^3\theta d\theta$$ $$A=\int(\cos\theta)^{1/2}\sin^3\theta d\theta$$ Since the factor of $\sin$ is odd, we would change $\sin$ in terms of $\cos$ and set aside 1 factor of $\sin$ for later substitution. $$A=\int(\cos\theta)^{1/2}\sin^2\theta\sin\theta d\theta$$ $$A=\int(\cos\theta)^{1/2}(1-\cos^2\theta)\sin\theta d\theta$$ Now let $u=\cos\theta$ We would have $du=-\sin\theta d\theta$, or $\sin\theta d\theta=-du$ That means $$A=-\int u^{1/2}(1-u^2) du$$ $$A=-\int (u^{1/2}-u^{5/2}) du$$ $$A=\int (u^{5/2}-u^{1/2}) du$$ $$A=\Bigg[\frac{u^{7/2}}{7/2}-\frac{u^{3/2}}{3/2}\Bigg]+C$$ $$A=\frac{2u^{7/2}}{7}-\frac{2u^{3/2}}{3}+C$$ $$A=\frac{2(\cos\theta)^{7/2}}{7}-\frac{2(\cos\theta)^{3/2}}{3}+C$$
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