## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 65

#### Answer

$3(1-(1/2)^{1/3}).$

#### Work Step by Step

Let $u=\cos \theta$, then $du=-\sin\theta d\theta$ and $u:1\to 1/2$ \begin{align*} \int_{0}^{\pi/3} \sin \theta \cos^{-2/3} \theta d \theta &=-\int_{1}^{1/2} u^{-2/3} du\\ &=-\frac{1}{1/3 }u^{1/3} |_1^{1/2}\\ &=3(1-(1/2)^{1/3}). \end{align*}

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