Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.2 Trigonometric Integrals - 7.2 Exercises: 13

Answer

$\frac{2}{7}(\cos\theta)^{7/2}-\frac{2}{3}(\cos\theta)^{3/2}+C$

Work Step by Step

$\int\sqrt{\cos\theta}\sin^3\theta\ d\theta$ $=\int\cos^{\frac{1}{2}}\theta\sin^2\theta\sin\theta\ d\theta$ $=\int\cos^{\frac{1}{2}}\theta(1-\cos^2\theta)\sin\theta\ d\theta$ Let $u=\cos\theta$. Then $du=-\sin\theta\ d\theta$, so $\sin\theta\ d\theta=-du$. $=\int u^{\frac{1}{2}}(1-u^2)*(-1)du$ $=-\int(u^{\frac{1}{2}}-u^{\frac{5}{2}})du$ $=-\left(\frac{u^{\frac{3}{2}}}{\frac{3}{2}}-\frac{u^{\frac{7}{2}}}{\frac{7}{2}}+C\right)$ $=\frac{2u^{\frac{7}{2}}}{7}-\frac{2u^{\frac{3}{2}}}{3}+C$ $=\boxed{\frac{2}{7}(\cos\theta)^{7/2}-\frac{2}{3}(\cos\theta)^{3/2}+C}$
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