Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 460: 48

Answer

$$\frac{\pi}{16}$$

Work Step by Step

Given $$\int_{0}^{1} t^{2} \sqrt{1-t^{2}} d t$$ Let $$t=\sin u\ \ \ \ \ \ \ \ \ dt =\cos udu $$ at $ t=0\to \ u=0$, at $ t=1\to \ u=\pi/2$ , so \begin{align*} \int_{0}^{1} t^{2} \sqrt{1-t^{2}} d t&=\int_{0}^{\pi/2} \sin^{2}u \sqrt{1-\sin^{2}u} \cos udu\\ &= \int_{0}^{\pi/2} \sin^{2}u \cos^2 udu\\ &=\frac{1}{4}\int_{0}^{\pi/2}(1-\cos 2u)(1+\cos 2u)du\\ &= \frac{1}{4}\int_{0}^{\pi/2}(1-\cos^2 2u)du \\ &= \frac{1}{4}\int_{0}^{\pi/2}\left(\frac{1}{2} +\frac{1}{2}\cos4u\right)du\\ &= \frac{1}{4}\left(\frac{1}{2}u +\frac{1}{8}\sin4u\right)\bigg|_{0}^{\pi/2}\\ &= \frac{\pi}{16} \end{align*}
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