Calculus 8th Edition

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

Chapter 7 - Techniques of Integration - 7.5 Strategy for Integration - 7.5 Exercises - Page 547: 16


$$\frac{\pi -2}{8}$$

Work Step by Step

Given $$\int_{0}^{\sqrt{2}/2}\frac{x^2}{\sqrt{1-x^2}}dx$$ Let $x=\sin u\ \ \Rightarrow \ \ dx =\cos udu$ , at $x=0\to u=0$ , at $x=\sqrt{2}/2\to u=\pi/4$ , then \begin{align*} \int_{0}^{\sqrt{2}/2}\frac{x^2}{\sqrt{1-x^2}}dx&=\int_{0}^{\pi/4}\frac{\sin^2u\cos udu}{\sqrt{1-\sin^2u}} \\ &=\int_{0}^{\pi/4} \sin^2u du \\ &=\frac{1}{2}\int_{0}^{\pi/4} (1-\cos2u) du\\ &=\frac{1}{2}\left[ u- \frac{1}{2}\sin 2u\right]_{0}^{\pi/4}\\ &=\frac{1}{2}\left[\frac{\pi}{4}- \frac{1}{2}\right]\\ &=\frac{\pi -2}{8} \end{align*}
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