Calculus (3rd Edition)

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

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 41


$ \frac{1}{5}(1-\frac{9\sqrt3}{32})$

Work Step by Step

Since $u=\cos x$, then $du=-\sin x dx$; moreover, when $x:0 \to \pi/6$, then $u:1 \to \sqrt 3/2$. Now, we have $$ \int_{0}^{\pi / 6} \sin x \cos ^{4} x d x=-\int_{1}^{\sqrt 3/ 2} u ^{4} d u\\ =-\frac{1}{5} u^5|_1^{\sqrt 3/2}= \frac{1}{5}(1-\frac{9\sqrt3}{32}).$$
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