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 485: 57


$\displaystyle \frac{1}{2}\pi-\frac{4}{3}$

Work Step by Step

$A=\displaystyle \int_{0}^{\pi}(\sin^{2}x-\sin^{3}x)dx$ $=\displaystyle \int_{0}^{\pi}[\frac{1}{2}(1-\cos 2x)-\sin x(1-\cos^{2}x)]dx$ $=\displaystyle \int_{0}^{\pi}\left(\frac{1}{2}-\frac{1}{2}\cos 2x\right)dx-\int_{0}^{\pi}\sin x(1-\cos^{2}x)]dx$ $=\displaystyle \left(\frac{1}{2}\pi-0\right)-(0-0)+\int_{1}^{-1}\left(1-u^{2}\right)du\quad $ $\left[\begin{array}{c}{u=\cos x}\\{du=-\sin xdx}\end{array}\right]$ $=\displaystyle \left(\frac{1}{2}\pi-0\right)-(0-0)+2\left[\frac{1}{3}u^{3}-u\right]_{0}^{1}$ $=\displaystyle \frac{1}{2}\pi+2\left(\frac{1}{3}-1\right)$ $=\displaystyle \frac{1}{2}\pi-\frac{4}{3}$
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