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: 6

Answer

$\frac{1}{2}\sin(t^2)-\frac{1}{3}\sin^3(t^2)+\frac{1}{10}\sin^5(t^2)$

Work Step by Step

$\int t \cos^5(t^2)dt$ Let $u=t^2$. Then $du=2tdt$, and $tdt=\frac{1}{2}\ du$. $=\int\cos^5(u)*\frac{1}{2}\ du$ Since the power of cosine is odd, save one factor of $\cos u$ and express the rest in terms of $\sin u$: $=\frac{1}{2}\int\cos^4 u\cos u\ du$ $=\frac{1}{2}\int(\cos^2 u)^2 \cos u\ du$ $=\frac{1}{2}\int(1-\sin^2 u)^2\cos u\ du$ Let $v=\sin u$. Then $dv=\cos u\ du$. $=\frac{1}{2}\int(1-v^2)^2\ dv$ $=\frac{1}{2}\int(1-2v^2+v^4)\ dv$ $=\frac{1}{2}(v-\frac{2}{3}v^3+\frac{1}{5}v^5)$ $=\frac{1}{2}v-\frac{1}{3}v^3+\frac{1}{10}v^5$ $=\frac{1}{2}\sin u-\frac{1}{3}\sin^3 u+\frac{1}{10}\sin^5 u$ $=\boxed{\frac{1}{2}\sin(t^2)-\frac{1}{3}\sin^3(t^2)+\frac{1}{10}\sin^5(t^2)}$
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